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5.11 Solving Optimization Problems

2 min read • february 15, 2024

In the last key topic , we began to take a look at optimization problems and even practiced a few. Let’s recap and do some more practice questions!

🔎 Understanding Optimization Problems

Optimization problems are a key aspect of real-world applications in calculus, and involve finding the maximum or minimum value of a function in applied contexts. These contexts can range from determining the dimensions for maximum volume to minimizing costs. The objective is to identify the optimal conditions that lead to an extreme value. 👾

📝 Optimization on the AP Calculus Exam

Optimization problems are presented in many formats on the AP Test— you may see them as part of multiple choice problems, and they are usually accompanied by lengthier context or story problems. You are, however, basically guaranteed to see them one way or another—which is why it’s so important to understand how to solve them!

🧩  How to Solve Optimization Problems

🧺 identify the objective function.

Begin by clearly defining the quantity you want to optimize. This is your objective function, often denoted by f ( x ) or f( y) . For instance, if you're a farmer looking to maximize your crop yield, your objective function might be P(x) for the total profit.

💥 Establish Constraints

Consider any constraints or limitations on the variables involved. Constraints could be in the form of limitations on resources, dimensions, or any other relevant factors. In our farming example, this could be the amount of land available or a budget constraint for purchasing seeds and fertilizer.

🖊️ Formulate the Optimization Equation

Create an equation that represents the quantity you want to optimize. This is the function you aim to maximize or minimize. If you're maximizing profit, your equation might involve revenue minus costs: P(x) = R(x) − C(x) .

🎯 Find Critical Points

Take the derivative of the objective function with respect to the variable of interest. Set the derivative equal to zero and solve for critical points. Remember, critical points are potential locations for maxima or minima.

🤔 Test Critical Points

Use the first or second derivative test to determine whether each critical point corresponds to a local maximum, minimum, or neither. This step ensures that you are pinpointing the desired extreme value.

🧠 Consider Endpoints

If the optimization problem involves a closed interval, evaluate the objective function at the endpoints as well. Include these results in your analysis.

📈 Optimization Practice Problems

Let’s put these steps into action and give two questions a try.

🏡  Problem 1: Maximizing Area of a Rectangular Garden

You have 100 meters of fencing to enclose a rectangular garden. What dimensions will maximize the enclosed area?

First, define your variables in terms of what was given to you.

• Let x be the width of the rectangle.
• The length, y , will be determined by the remaining fencing: 2 x + 2 y = 100 2x+2y=100 2 x + 2 y = 100 or y = 50 − x . y=50 -x. y = 50 − x .

Since we’re asked to maximize the area, we can define the equations with the area formula of a rectangle. The area, A , of the rectangle is given by

We have our equations! Now we can find our critical points . Take the derivative of A ( x ) A(x) A ( x ) and set it equal to zero:

Test this critical point using the second derivative test. This will confirm that x = 25 x=25 x = 25 is a maximum.

Because A ′ ′ ( x ) A''(x) A ′′ ( x ) is a negative value, this confirms that x = 25 x = 25 x = 25 is a maximum.

Last but not least, interpret results!

The dimensions that maximize the area are x = 25   m x = 25 \ m x = 25   m (width) and y = 25   m y = 25 \ m y = 25   m (length). Good job! 👏

📦  Problem 2: Maximize the Size of a Can

A cylindrical can is to be made to contain 1000 π 1000\pi 1000 π cubic centimeters of liquid. Find the dimensions (radius and height) of the can that minimize the amount of material needed to manufacture the can.

Let’s again define our variables. Let r represent the radius of the cylinder and h represent the height of the cylinder.

Now we can define and simplify our equations. The quantity to be optimized in this problem is the surface area of the cylinder, which is the sum of the lateral surface area and the area of the two circular bases. The surface area A is given by:

The problem states that the can must contain 1000 π 1000\pi 1000 π cubic centimeters of liquid. The volume V of a cylinder is given by:

Since V = 1000 π = π r 2 h V = 1000\pi = \pi r^2h V = 1000 π = π r 2 h , we have r 2 h = 1000 , r^2h = 1000, r 2 h = 1000 , which serves as the constraint equation.

Solve the constraint equation for h:

Then, substitute h into the equation for surface area, A A A :

Simplify this equation:

Find the critical points! Take the derivative of A with respect to r and set it equal to zero to find critical points:

Check the values of r at the endpoints of the feasible interval (in this case, r cannot be negative, so only consider positive values).

Determine the minimum values. Evaluate the area function A at the critical point and endpoints, and determine which one gives the minimum value. In this case, r = 500 3 r = \sqrt[3]{500} r = 3 500 ​ yields the minimum surface area.

Use the expression h = 1000 r 2 h = \frac{1000}{r^2} h = r 2 1000 ​ to find the corresponding value of the height h . The answer is approximately h = 63.

You’re almost there! State the conclusion.

The dimensions that minimize the amount of material needed to manufacture the can are r = 500 3 r = \sqrt[3]{500} r = 3 500 ​ centimeters and h = 1000 ( 500 3 ) 2 h = \frac{1000}{(\sqrt[3]{500})^2} h = ( 3 500 ​ ) 2 1000 ​ , or approximately 63 centimeters.

🏆 Tips for Success

You made it to the end of this guide! Here are some tips for success:

• 💡  Clearly Define Variables: Ensure a clear understanding of the meaning of each variable in the problem.
• 📈  Graphical Insight: Consider graphing the function to visualize critical points and endpoints.
• 🧠  Units Matter: Pay attention to units in real-world problems. Ensure your final answer makes sense in the given context.

Happy optimizing!

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• Applied Mathematics

Section 5.6 - Optimization and Modeling

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5.5: Optimization

• Last updated
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• Page ID 108809

• Shana Calaway, Dale Hoffman, & David Lippman
• Shoreline College, Bellevue College & Pierce College via The OpenTextBookStore

The partial derivatives tell us something about where a surface has local maxima and minima. Remember that even in the one-variable cases, there were critical points which were neither maxima nor minima – this is also true for functions of many variables. In fact, as you might expect, the situation is even more complicated.

Second Derivatives

When you find a partial derivative of a function of two variables, you get another function of two variables – you can take its partial derivatives, too. We've done this before, in the one-variable setting. In the one-variable setting, the second derivative gave information about how the graph was curved. In the two-variable setting, the second partial derivatives give some information about how the surface is curved, as you travel on cross-sections – but that's not very complete information about the entire surface.

Imagine that you have a surface that's ruffled around a point, like what happens near a button on an overstuffed sofa, or a pinched piece of fabric, or the wrinkly skin near your thumb when you make a fist. Right at that point, every direction you move, something different will happen – it might increase, decrease, curve up, curve down… A simple phrase like concave up or concave down can't describe all the things that can happen on a surface.

Surprisingly enough, though, there is still a second derivative test that can help you decide if a point is a local max or min or neither, so we still do want to find second derivatives.

Second Partial Derivatives

Suppose \( f(x,y) \) is a function of two variables. Then it has four second partial derivatives : \[ \begin{align*} f_{xx} & = \frac{\partial}{\partial x}\left(f_x\right)=\left(f_x\right)_x & f_{xy} & = \frac{\partial}{\partial y}\left(f_x\right)=\left(f_x\right)_y\\ f_{yx} & = \frac{\partial}{\partial x}\left(f_y\right)=\left(f_y\right)_x & f_{yy} & = \frac{\partial}{\partial y}\left(f_y\right)=\left(f_y\right)_y \end{align*} \nonumber \] \( f_{xy} \) and \( f_{yx} \) are called the mixed (second) partial derivatives of \(f\) .

Leibniz notation for the second partial derivatives is a bit confusing, and we won't use it as often: \[ \begin{align*} f_{xx} & = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial^2 f}{\partial x^2} & f_{xy} & = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial^2 f}{\partial y \partial x}\\ f_{yx} & = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial^2 f}{\partial x \partial y} & f_{yy} & = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial^2 f}{\partial y^2} \end{align*} \nonumber \]

Notice that the order of the variables for the mixed partials goes from right to left in the Leibniz notation instead of left to right.

Example \(\PageIndex{1}\)

Find all four partial derivatives of \( f(x,y)=x^2-4xy+4y^2 \).

We have to start by finding the (first) partial derivatives: \[ \begin{align*} f_x(x,y) & = 2x-4y \\ f_y(x,y) & = -4x+8y \end{align*} \nonumber \]

Now we’re ready to take the second partial derivatives: \[ \begin{align*} f_{xx}(x,y) & = \frac{\partial}{\partial x}(2x-4y)=2 \\ f_{xy}(x,y) & = \frac{\partial}{\partial y}(2x-4y)=-4 \\ f_{yx}(x,y) & = \frac{\partial}{\partial x}(-4x+8y)=-4 \\ f_{yy}(x,y) & = \frac{\partial}{\partial y}(-4x+8y)=8 \end{align*} \nonumber \]

You might have noticed that the two mixed partial derivatives were equal in this last example. It turns out that it's not a coincidence – it's a theorem!

Mixed Partial Derivative Theorem

If \( f \), \( f_x \), \( f_y \), \( f_{xy} \), and \( f_{yx} \) are all continuous (no breaks in their graphs), then \[ f_{xy}=f_{yx}. \nonumber \]

In fact, as long as \(f\) and all its appropriate partial derivatives are continuous, the mixed partials are equal even if they are of higher order, and even if the function has more than two variables.

This theorem means that the confusing Leibniz notation for second derivatives is not a big problem – in almost every situation the mixed partials are equal, so the order in which we compute them doesn't matter.

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Example \(\PageIndex{2}\)

Find \( \frac{\partial^2 f}{\partial x \partial y} \) for \( f(x,y)=\frac{e^{x+y}}{y^3+y}+y\ln(y) \).

We already found the first partial derivatives in an earlier example: \[ \begin{align*} \frac{\partial f}{\partial x} & = \frac{1}{y^3+y}e^{x+y} \\ \frac{\partial f}{\partial y} & = \frac{\left( e^{x+y}(1) \right)\left( y^3+y \right)-\left( e^{x+y} \right)\left( 3y^2+1 \right)}{\left( y^3+y \right)^2}+\left( 1 \right)\left( \ln(y) \right)+\left( y \right)\left( \frac{1}{y} \right) \end{align*} \nonumber \]

Now we need to find the mixed partial derivative. The theorem says that \( \frac{\partial f^2}{\partial x \partial y}=\frac{\partial f^2}{\partial y \partial x} \), so it doesn't matter whether we find the partial derivative of \( \frac{\partial f}{\partial x} \) with respect to \(y\) or the partial derivative of \( \frac{\partial f}{\partial y} \) with respect to \(x\). Which would you rather do?

It looks like it will be easier to compute the mixed partial by finding the partial derivative of \( \frac{\partial f}{\partial x}= \frac{1}{y^3+y}e^{x+y} \) with respect to \(y\) – it still looks messy, but it looks less messy: \[ \frac{\partial f^2}{\partial y \partial x}= \frac{\partial}{\partial y}\left(\frac{1}{y^3+y}e^{x+y}\right)=\frac{\left(e^{x+y}\right)\left(y^3+y\right)-\left(e^{x+y}\right)\left(3y^2+1\right)}{\left(y^3+y\right)^2} \nonumber \]

If we had decided to do this the other way, we'd end up in the same place. Eventually…

Local Maxima, Local Minima, and Saddle Points

Let's briefly review max-min problems in one variable.

A local max is a point on a curve that is higher than all the nearby points. A local min is lower than all the nearby points. We know that local max or min can only occur at critical points, where the derivative is zero or undefined. But we also know that not all critical points are max or min, so we also need to test them, with the First Derivative or Second Derivative Test.

The situation with a function of two variables is much the same. Just as in the one-variable case, the first step is to find critical points, places where both the partial derivatives are either zero or undefined.

Definition: Local Maximum and Minimum

• \(f\) has a local maximum at \((a, b)\) if \(f(a, b) \geq f(x, y)\) for all points \((x, y)\) near \((a, b)\).
• \(f\) has a local minimum at \((a, b)\) if \(f(a, b) \leq f(x, y)\) for all points \((x, y)\) near \((a, b)\).

A critical point for a function \(f(x, y)\) is a point \((x, y)\) (or \((x, y, f(x, y))\)) where both the following are true:

• \( f_x=0 \) or is undefined, and
• \( f_y=0 \) or is undefined.

Just as in the one-variable case, a local max or min of \(f\) can only occur at a critical point.

Just as in the one-variable setting, not all critical points are local max or min. For a function of two variables, the critical point could be a local max, local min, or a saddle point.

A point on a surface is a local maximum if it's higher than all the points nearby; a point is a local minimum if it's lower than all the points nearby.

A saddle point is a point on a surface that is a minimum along some paths and a maximum along some others. It's called this because it's shaped a bit like a saddle you might use to ride a horse. You can see a saddle point by making a fist – between the knuckles of your index and middle fingers, you can see a place that is a minimum as you go across your knuckles, but a maximum as you go along your hand toward your fingers.

Here is a picture of a saddle point from a few different angles. This is the surface \( f(x,y)=5x^2-3y^2+10 \), and there is a saddle point above the origin. The lines show what the surface looks like above the \(x\)- and \(y\)- axes. Notice how the point above the origin, where the lines cross, is a local minimum in one direction, but a local maximum in the other direction.

Second Derivative Test

Just as in the one-variable case, we'll need a way to test critical points to see whether they are local max or min. There is a second derivative test for functions of two variables that can help, but, just as in the one-variable case, it won't always give an answer.

The Second Derivative Test for Functions of Two Variables

• Find all critical points of \(f(x,y)\).
• If \( f_{xx}\gt 0 \), then \(f\) has a local minimum at the critical point.
• If \( f_{xx}\lt 0 \), then \(f\) has a local maximum at the critical point.
• If \(D \lt 0\) then \(f\) has a saddle point at the critical point.
• If \(D = 0\), there could be a local max, local min, or neither (i.e., the test in inconclusive).

Example \(\PageIndex{3}\)

Find all local maxima, minima, and saddle points for the function \[ f(x,y)=x^3+y^3+3x^2-3y^2-8. \nonumber \]

First we need the partial derivatives: \( f_x=3x^2+6x \) and \( f_y=3y^2-6y \).

Critical points are the places where both of these are zero (neither is ever undefined): \( f_x=3x^2+6x=3x(x+2)=0 \) when \(x = 0\) or when \(x = -2\). \( f_y=3y^2-6y=3y(y-2)=0 \) when \(y = 0\) or when \(y = 2\).

Putting these together, we get four critical points: (0, 0), (-2, 0), (0, 2), and (-2, 2).

Now to classify them, we’ll use the Second Derivative Test. We’ll need all the second partial derivatives: \[f_{xx}=6x+6,\ f_{yy}=6y-6,\ f_{xy}=f_{yx}=0.\nonumber \]

Then \[ D(x,y)=(6x+6)(6y-6)-(0)(0)=(6x+6)(6y-6). \nonumber \]

Now look at each critical point in turn:

• At (0, 0): \( D(0,0)=(6(0)+6)(6(0)-6)=(6)(-6)=-36 \lt 0 \), so there is a saddle point at (0, 0).
• At (-2, 0): \( D(-2,0)=(6(-2)+6)(6(0)-6)=(-6)(-6)=36 \gt 0 \) and \( f_{xx}(-2,0)=6(-2)+6=-6 \lt 0 \), so there is a local maximum at (-2, 0).
• At (0, 2): \( D(0,2)=(6(0)+6)(6(2)-6)=(6)(6)=36 \gt 0 \) and \( f_{xx}(0,2)=6(0)+6=6 \gt 0 \), so there is a local minimum at (0, 2).
• At (-2, 2): \( D(-2,2)=(6(-2)+6)(6(2)-6)=(-6)(6)=-36 \lt 0 \), so there is another saddle point at (-2, 2).

Example \(\PageIndex{4}\)

Find all local maxima, minima, and saddle points for the function \[ z=9x^3+\frac{y^3}{3}-4xy. \nonumber \]

We’ll need all the partial derivatives and second partial derivatives, so let’s compute them all first: \[ \begin{align*} z_x & = 27x^2-4y,\quad z_y= y^2-4x,\\ z_{xx} & = 54x,\quad z_{xy}=z_{yx}= -4,\quad z_{y}= 2y. \end{align*} \nonumber \]

Now to find the critical points: We need both \( z_x \) and \( z_y \) to be zero (neither is ever undefined), so we need to solve this set of equations simultaneously: \[ \begin{align*} z_x & = 27x^2-4y=0 \\ z_y & = y^2-4x=0 \end{align*} \nonumber \]

Perhaps it's been a while since you solved systems of equations. One solution method is the substitution method – solve one equation for one variable and substitute into the other equation: \[ \left.\begin{align*} 27x^2-4y & = 0 \\ y^2-4x & = 0 \end{align*}\right\} \to \text{Solve \( y^2-4x=0 \) for \( x=\frac{y^2}{4} \) \( \dots \)} \nonumber \] …then substitute into the other equation: \[ \begin{align*} 27\left(\frac{y^2}{4}\right)^2-4y & = 0 \\ \frac{27}{16}y^4-y & = 0 \end{align*} \nonumber \]

Now we have just one equation in one variable to solve. Factoring out a \(y\) gives \[ y\left(\frac{27}{16}y^3-1\right)=0, \nonumber \] so \( y=0 \) or \( \frac{27}{16}y^3-1=0 \), giving \( y=\sqrt[3]{\frac{1}{27/16}}=\frac{\sqrt[3]{4}}{3} \).

Plugging back in to the equation \( x=\frac{y^2}{4} \) to find \(x\) gives us the two critical points: (0,0) and \( \left(\frac{4}{9},\frac{4}{3}\right) \).

Now to test them. First compute \[ \begin{align*} D(x,y) & = (f_{xx})(f_{yy})-(f_{xy})(f_{yx}) \\ & = (54x)(2y)-(-4)(-4) \\ & = 108xy-16 \end{align*} \nonumber \] Then evaluate \( D \) at the two critical points:

• At (0,0): \(D(0,0) = -16 \lt 0\), so there is a saddle point at (0, 0).
• At \( \left(\frac{4}{9},\frac{\sqrt[3]{4}}{3}\right) \): \(D\left(\frac{4}{9},\frac{\sqrt[3]{4}}{3}\right) =16(\sqrt[3]{4}-1) \gt 0\), and \( f_{xx}\left(\frac{4}{9},\frac{\sqrt[3]{4}}{3}\right) \gt 0 \), so there is a local minimum at \( \left(\frac{4}{9},\frac{\sqrt[3]{4}}{3}\right) \).

Applied Optimization

Example \(\pageindex{5}\).

A company makes two products. The demand equations for the two products are given below. \(p_1\), \(p_2\), \(q_1\),and \(q_2\) are the prices and quantities for Products 1 and 2. \[ \begin{align*} q_1 & = 200-3p_1-p_2 \\ q_2 & = 150-p_1-2p_2 \end{align*} \nonumber \]

Find the price the company should charge for each product in order to maximize total revenue. What is that maximum revenue?

Revenue is still price\( \cdot \)quantity. If we're selling two products, the total revenue will be the sum of the revenues from the two products: \[ \begin{align*} R(p_1,p_2) & = p_1q_1+p_2q_2 \\ & = p_1(200-3p_1-p_2)+p_2(150-p_1-2p_2) \\ & = 200p_1-3p_1^2-2p_1p_2+150p_2-2p_2^2 \end{align*} \nonumber \]

This is a function of two variables, the two prices, and we need to optimize it (just as in the previous examples). First we find critical points. The notation here gets a bit hard to look at, but hang in there – this is the same stuff we've done before. \[ R_{p_1}=200-6p_1-2p_2 \ \text{ and } \ R_{p_2}=150-2p_1-4p_2. \nonumber \]

Solving these simultaneously gives the one critical point \( (p_1, p_2)=(25,25) \)

To confirm that this gives maximum revenue, we need to use the Second Derivative Test. Find all the second derivatives: \[ R_{p_1 p_2}=-6,\ R_{p_2 p_2}=-4,\ \text{ and }\ R_{p_1 p_2}=R_{p_2 p_1}=-2. \nonumber \]

So \( D(25,25)=(-6)(-4)-(-2)(-2) \gt 0 \) and \( R_{p_1 p_2}(25,25) \lt 0 \), so this really is a local maximum.

Thus, to maximize revenue the company should charge $25 per unit for both products. This will yield a maximum revenue of $4375.

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Chapter 5 – Quadratics, Polynomials and Rational Expressions

Topic 5.6 – Optimization Using Parabolas

Slideshow: Full –  4 per page – 9 per page

Algebra Copyright © 2022 by Mike Weimerskirch and the University of Minnesota Board of Regents is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License , except where otherwise noted.

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Chapter 5.7: Applied Optimization Problems

Learning objectives.

• Set up and solve optimization problems in several applied fields.

One common application of calculus is calculating the minimum or maximum value of a function. For example, companies often want to minimize production costs or maximize revenue. In manufacturing, it is often desirable to minimize the amount of material used to package a product with a certain volume. In this section, we show how to set up these types of minimization and maximization problems and solve them by using the tools developed in this chapter.

Solving Optimization Problems over a Closed, Bounded Interval

The basic idea of the optimization problems that follow is the same. We have a particular quantity that we are interested in maximizing or minimizing. However, we also have some auxiliary condition that needs to be satisfied. For example, in (Figure) , we are interested in maximizing the area of a rectangular garden. Certainly, if we keep making the side lengths of the garden larger, the area will continue to become larger. However, what if we have some restriction on how much fencing we can use for the perimeter? In this case, we cannot make the garden as large as we like. Let’s look at how we can maximize the area of a rectangle subject to some constraint on the perimeter.

Maximizing the Area of a Garden

A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides ( (Figure) ). Given 100 ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?

Determine the maximum area if we want to make the same rectangular garden as in (Figure) , but we have 200 ft of fencing.

Now let’s look at a general strategy for solving optimization problems similar to (Figure) .

Problem-Solving Strategy: Solving Optimization Problems

• Introduce all variables. If applicable, draw a figure and label all variables.
• Determine which quantity is to be maximized or minimized, and for what range of values of the other variables (if this can be determined at this time).
• Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may involve more than one variable.
• Write any equations relating the independent variables in the formula from step 3. Use these equations to write the quantity to be maximized or minimized as a function of one variable.
• Identify the domain of consideration for the function in step 4 based on the physical problem to be solved.
• Locate the maximum or minimum value of the function from step 4. This step typically involves looking for critical points and evaluating a function at endpoints.

Now let’s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.

Maximizing the Volume of a Box

An open-top box is to be made from a 24 in. by 36 in. piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?

To find the critical points, we need to solve the equation

Dividing both sides of this equation by 12, the problem simplifies to solving the equation

Using the quadratic formula, we find that the critical points are

Watch a video about optimizing the volume of a box.

Minimizing Travel Time

and the time spent swimming is

Therefore, the total time spent traveling is

which implies

Therefore, the possibilities for critical points are

In business, companies are interested in maximizing revenue . In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let’s use these data to determine the price the company should charge to maximize the amount of money it brings in.

Maximizing Revenue

Maximizing the Area of an Inscribed Rectangle

A rectangle is to be inscribed in the ellipse

What should the dimensions of the rectangle be to maximize its area? What is the maximum area?

Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded

In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domain is neither closed nor bounded.

In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area.

Minimizing Surface Area

A rectangular box with a square base, an open top, and a volume of 216 in. 3 is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?

Key Concepts

• To solve an optimization problem, begin by drawing a picture and introducing variables.
• Find an equation relating the variables.
• Find a function of one variable to describe the quantity that is to be minimized or maximized.
• Look for critical points to locate local extrema.

For the following exercises, answer by proof, counterexample, or explanation.

1. When you find the maximum for an optimization problem, why do you need to check the sign of the derivative around the critical points?

The critical points can be the minima, maxima, or neither.

2. Why do you need to check the endpoints for optimization problems?

For the following exercises, set up and evaluate each optimization problem.

7. Find the positive integer that minimizes the sum of the number and its reciprocal.

8. Find two positive integers such that their sum is 10, and minimize and maximize the sum of their squares.

For the following exercises, consider the construction of a pen to enclose an area.

16. You can run at a speed of 6 mph and swim at a speed of 3 mph and are located on the shore, 4 miles east of an island that is 1 mile north of the shoreline. How far should you run west to minimize the time needed to reach the island?

21. Find the cheapest driving speed.

22. Find the profit function for the number of pizzas. How many pizzas gives the largest profit per pizza?

For the following exercises, draw the given optimization problem and solve.

31. Find the volume of the largest right circular cylinder that fits in a sphere of radius 1.

32. Find the volume of the largest right cone that fits in a sphere of radius 1.

For the following exercises, consider the points on the given graphs. Use a calculator to graph the functions.

For the following exercises, set up, but do not evaluate, each optimization problem.

42. You have a garden row of 20 watermelon plants that produce an average of 30 watermelons apiece. For any additional watermelon plants planted, the output per watermelon plant drops by one watermelon. How many extra watermelon plants should you plant?

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