# Optimization Problems Homework Mastermathmentor Calculus

## AB Calculus Manual (Revised 1/2016)

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One of the most challenging aspects of calculus is optimization. Many AP Calculus students struggle with optimization problems because they require a bit more critical thinking than a normal problem. Reading this article will give you all the tools you need to solve optimization problems, including some examples that I will walk you through. Together, we will beat all of your fears and confusion. Let’s get started.

First, what is optimization? Optimization is when we are looking for the extrema of a function. Extrema are the maximum or minimum values. We can have absolute extrema and local extrema. Local extrema are the peaks and troughs in an equation. Absolute extrema are the overall maximum values or the overall minimum values. Absolute extrema can be within the function or they can be at the ends of the interval we are searching for the extrema on.

There are many different types of optimization problems. We could be optimizing volume, area, distance, length, and many other quantities. These are just some common, simple examples.

The types of optimization problems that we will be covering in this article involve something called a constraint. A constraint can be an equation, and a constraint is always true in the concept of the problem. These problems become difficult in AP Calculus because students can become confused about which equation we are trying to optimize and which equation represents the constraint. The best way to prevent this confusion is to read the problem very carefully, draw picture representations of whatever you are trying to optimize, and label your equation and your constraint. Always do this first before solving any problem.

Let’s work through several examples of optimization problems in order to gain a better understanding of the concept. In each example, pay attention to the precise wording of the problem. Reading as many examples as you can and becoming more acquainted with the structure of these problems will help you get better at interpreting them.

#### Example 1

We want to create a box with an open top and square base with a surface area of 300 square inches. What height will produce a box with the maximum possible volume?

**Step 1:** Identify the equation we want to maximize.

Reading the problem, we see that we want to maximize the volume, but solve for the height of the box. The equation for the volume of a cube is:

V=x ^2h

In this equation, the x represents the two side measurements of the box and h represents the height of the box.

**Step 2:** Identify the constraint equation.

When working these optimization problems, it is important to remember that we always need two equations. Why? The reason will become clear later. Reading the problem, we see that the surface area of the box is a constant 300 square inches. This is the constraint. We can turn this into an equation of the surface area:

SA= x^2+4xh=300

The x^2 term represents the square base of the box and the 4xh term represents the four sides of the box. We couldn’t employ the normal surface area equation for a cube to this example because the top of the box is open.

**Step 3:** Eliminate a variable.

Looking at both of our equations, we see that each is composed of two variables. We know that in order to solve this equation, we must somehow eliminate one variable. Because we have two equations, (I told you I’d explain the importance) we can solve the constraint equation to be in terms of one of the variables and plug that equation into our volume equation – the equation we are trying to maximize. Our first instinct would be to solve the constraint equation for in order to eliminate it from the volume equations (since our goal is to find the optimal height) but we see that solving for x would be relatively more difficult than solving for h. So, for our own ease, we will solve for h and eliminate it for now. Don’t worry, we will solve for it at the end of the problem.

h=\frac{300-x^2}{4x}

We can then plug this equation into the volume equation in place of height.

V=\frac{x\left(300-x^2\right)}{4}

**Step 4:** Determine the bounds of the problem.

There are two extremes to this problem:

1. The box is all height and no side length (x=0).

2. The box is all side length and there is no height, (solve the height equation for when h=0).

0=\frac{300-x^2}{4x}

x=\sqrt{300}=17.32

Therefore, the bounds that we are looking for the maximum on are from x=0 to x=17.32. This is useful information because we cannot go outside the scope of our constraint with these bounds and possibly solve for the wrong maximum.

**Step 5:** Find the critical numbers.

Critical numbers are the values of x where the function has a slope of zero. These critical numbers will tell us where it is possible for extrema to occur. In this problem we are trying to find the maximum volume. We can find these critical numbers by taking the derivative of the volume equation and setting it equal to zero.

\frac{dV}{dx}= 75-\frac{3}{4}x^2=0

x=10

At x=10, either a maximum or a minimum occurs.

**Step 6:** Find the maximum volume.

We now test each of our bounds and our critical numbers in the volume equation to see which one has the maximum value.

For x=0, V=0.

For x=10, V=500.

For x=17.32, V=0.0044

We see from this example that the maximum value of volume is 500 cubic inches.

**Step 7:** Solve for height.

Because we know the side length value that corresponds to the maximum volume, we can solve for the corresponding height using the constraint equation we rearranged in Step 3.

h=\frac{300-{10}^2}{4\left(10\right)}

From this example, we see the basic algorithm for solving these problems. We identify the equations, solve for a variable, use derivatives to find the critical numbers, and then test to find the maximum or minimum value. The most difficult part of any optimization problem is interpreting the problem statement. Let’s see another example.

#### Example 2

Find the point on the curve y= x^2 that is closest to the point (1,5).

At the onset of this problem we realize that we want to minimize the distance between the given curve and a specific point on our coordinate system.

**Step 1:** Identify the equation we want to minimize.

You would think that we need to minimize the curve equation as a first guess, but this guess is not correct. We want to minimize the distance between a point on the curve and the given point. Therefore, we want to minimize the distance equation:

d=\sqrt{{\left(x-1\right)}^2+{\left(y-5\right)}^2}

**Step 2:** Identify the constraint equation.

The only equation that we know is constantly true in this scenario is the equation of the curve; therefore, it becomes our constraint equation.

y= x^2

**Step 3:** Eliminate a variable.

This step is easier in this example than the last one. We see that the distance equation has both x and y. Our constraint equation is already solved for y! Let’s just plug it in:

d=\sqrt{{\left(x-1\right)}^2+{\left(x^2-5\right)}^2}

**Step 4:** Determine the bounds of the problem.

In this specific example there are no specific bounds from the constraint equation that we need to test for. Therefore, we define the bounds at negative infinity and positive infinity.

**Step 5:** Find the critical numbers.

Remember that critical numbers are found by setting the derivative of our equation equal to zero!

\frac{dd}{dx}=\frac{\left(2x^3-9x-1\right)}{\sqrt{x^4- 9x^2-2x+26}}=0

x=2.1748

**Step 6:** Find the minimum distance.

Plug the bounds and the critical values into the distance formula we derived and see which number has the minimum distance. We won’t try our bounds in this case because they aren’t finite numbers.

For x=2.1748, d=1.2055

Since we only have one critical number, we only have one distance. This is the minimum distance then, by default.

**Step 7:** Solve for the y value

We just plug the x value into the constraint equation.

y= {\left(2.1748\right)}^2

y=4.7298

Now we are starting to really understand these problems. Let’s look at one more example.

#### Example 3

A high school student wants to draw a rectangle into a semicircle of radius 5. If one side must be on the semicircle’s diameter, what is the area of the largest rectangle the student can draw?

First, we notice that we are trying to maximize the area of a rectangle that is inside a circle.

**Step 1:** Identify the equation we want to minimize.

As we said before, we want to minimize the area of a rectangle. The area of a square is given by the equation:

A=2xy

In this equation, x is half the width of the base of the rectangle.

**Step 2:** Identify the constraint equation.

The area of the rectangle is constrained by the bounds of the semicircle. So the equation of the semicircle is the constraint.

y=\sqrt{5^2-x^2}

**Step 3:** Eliminate a variable.

We see that our area equation has two variables, x and y. We will eliminate the y variable by plugging the semicircle equation in for y:

A=2x\sqrt{5^2-x^2}

**Step 4:** Determine the bounds of the problem.

We realize that the base of the semicircle can only be as long as the diameter of the semicircle since one side is resting on the diameter. Therefore, can only be as long as the radius of the semicircle, which is 5. The bounds, then, are from x=0 to x=5.

**Step 5:** Find the critical numbers.

We find the critical numbers of the area equation by taking the derivative of the area function. We then set the derivative equal to zero and solve for the corresponding x values.

\frac{dA}{dx}=2\sqrt{25-x^2}-\frac{2x^2}{\sqrt{25-x^2}}=0

x=\frac{-5\sqrt{2}}{2}

x=\frac{5\sqrt{2}}{2}

We see that the negative critical number is not within the bounds we set in step 4. Therefore, we throw it out.

**Step 6:** Find the maximum area.

We will plug each of the bounds and valid critical numbers into the area equation we derived to figure out which value yields the maximum rectangle area.

For x=0, A=0.

For x=\frac{5\sqrt{2}}{2}, A=25

For x=5, A=0

We see from this analysis that the maximum area is achieved at our critical number. We do not need to proceed to step 7 since we have already calculated the maximum area of the rectangle, 25.

When working these problems, it is important to always remember these steps that I have demonstrated in these example problems. Even though it is not explained in this article, it is also important to be able to take derivatives of functions. Some functions are more tricky, so be sure to review derivatives when studying for the AP Calculus test.

Another important thing to note is the ability to craft geometric equations. I bet you never thought your geometry class would ever become useful, but things change. It is very important to be able to recall the area, surface area, and volume of all the common shapes, including squares, rectangles, circles, triangles, spheres, cubes, prisms, and pyramids. It is also important to know where these equations came from in cases like the first example, where a shape may be missing a side. You may end up deriving your own equations!

The most important way to prepare for optimization problems on the AP Calculus exam is to practice. I know I’ve already mentioned that in this article, but practice is extremely important. Go back and work the homework problems your teacher gave you. Work these examples without looking at their solutions. Look up additional problems online. There are plenty of resources out there for you to harness.

Optimization is one of the most challenging parts of AP Calculus. If you can conquer these problems, you can do anything. I wish you the best of luck with your AP Calculus review. Remember to practice. Do you know something that we haven’t covered in this article? Let us know!

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