Becka and Becca, Math SI’s 😛

This week’s blog is going to focus on optimization. Since we are math tutors for both calculus and pre-calculus, we are going to show you how to solve this problem first with precalculus. Then, for students who are already taking calculus, we are going to provide you with an alternative route for achieving the same solution! Yay!!

First, let’s start by defining what optimization is – finding the most advantageous solution to a problem. Examples of this would be looking to get the most out of a product or finding a method which creates the least amount of waste. These solutions are the maximums and minimums.

Let’s look at an example!

A rain gutter is formed by bending up the sides of a 30-in. wide rectangular metal sheet.

1. Find a function that models the cross-sectional area of the gutter in terms of x.
2. Find the value of x that maximizes the cross-sectional area of the gutter.
3. What is the maximum cross-sectional area for the gutter?

Where do we start? Let’s start with what we are given (what are our knowns?):

We know that the sheet used to create the gutter has to have a total width of 30 inches. Then we will fold up the sides to create the gutter. We also know that, by convention, each of the sides that are folded up have to be equal. We can then generate a picture to better visualize our problem:

Finally, we know that we want to optimize the area created by folding up the two sides of length x. (See the figure above)

So what do they mean by the cross-sectional area? When we look at a cross-section, we are cutting a slice of the gutter to look at a 2D version of it. In other words, what we would see if we were to look at the frontal view of the gutter (reference the picture above).

According to part (a) of our problem, we need to find a function that models the cross-sectional area of the gutter. Based on the picture, we are just finding the area of this rectangle, which can be denoted by the function

f(x)=base x height

f(x)=(30-2x)(x)

f(x)=30x – 2x^2

Now that we have part (a) answered, we can go to part (b) of the problem, which asks us to find the x that optimizes the cross-sectional area. This means we need to find where the maximum occurs which is at the vertex of our function. In order to find our vertex, we need to put our function into standard form. The general formula for standard form is:

f(x)=a(x-h)^2+k

Where the point(h,k) is our vertex. We use completing the square to get our function in this form.

Calculus Method

We can also answer parts (b) and (c) of this question using calculus! Instead of putting the equation into standard form, we can take the derivative. The derivative of our function is

f'(x)=30-4x

But, how does this relate to the maximum of the function? The derivative of the function tells us the slope of the function. In other words, when the slope equals 0 and the function goes from a positive to a negative slope at that x value. This means we have found our maximum.

This matches the answer we got by using precalculus, so we know that it’s correct!

Here’s a problem to try on your own!

A farmer has 400 feet of fencing and wants to enclose a pasture next to a river for his cows. If the farmer puts fencing on all sides except against the river, what is the largest area he can enclose?

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The largest area would be 200 ft^2.

If you didn’t know how to start this problem or how we got our answer, come to CAPS! We would be happy help!