Mr B's Math Blog

Sunday, December 20, 2009

Math 350: The dreaded rectangle problem

Problem 16 on the final exam appears to have been no one's idea of a good time. It's one of those multi-step problems whose solution depends on a successful start, and not too many people achieved that. For future reference, here's a step-by-step solution that might be instructive to people who want to be prepared when they see this sort of thing again (if ever).

First of all, here's the statement of the problem:

Obadiah has an oddly shaped plot of land that is bounded by two straight lines and a parabola. He has set up a coordinate system so that his plot of land is bounded by the x axis, the line x = 4, and the curve y = x^1/2. (See the figure.) He wants to plant a rectangular patch of grass on his land and he wants it to have the greatest possible area. Assuming that the units are expressed in yards, what are the dimensions and the area of the rectangle with maximum area? (Work out the following steps in answering this question.)

The problem included three steps, the first of which said, “Use the figure to express the length and width of the rectangle in terms of x and/or y.” We can do this if we know how to read distances from a rectangular coordinate system. Here's one thing to keep in mind:

The value of x is a variable and the point (x, y) can be anywhere on the curve y = x^1/2. We do not try to read actual numbers from the graph when we are trying to express things in terms of x and y. Any one of the illustrations immediately below could have accompanied the exam problem (we actually used the middle one, which caused some people to guess that x was supposed to be 1.75—but nope). Our job is to find the x and y that will give us the maximum area:

We also want to take advantage of this key fact: The coordinates of a point in a rectangular coordinate system automatically give us two distances we can use: the horizontal and vertical distances (respectively) of the point from the origin (0, 0):

We should see right away that x and y do not give the length and width of the rectangle. In the past we saw problems where A = xy was the formula for the area, but this one is not quite that simple. Yes, y gives the width of the rectangle, but x is not its length. What is the length? Consider this:

We see that the entire horizontal distance from the origin to the right edge of the rectangle is 4. The horizontal distance to the left edge of the rectangle is x. The length of the rectangle is the difference of these two differences. That is, since

x + L = 4

we have

Length = L = 4 − x.

We already saw that Width = y, so

Area = (Length)(Width) = (4 − x)(y).

The second step of Problem 16 said, “The area of the rectangle is given by length times width. Write the area as a function of x alone and find its derivative.” We know that y = x^1/2, so we can substitute this into the formula for area:

Area = (4 − x)(x^1/2).

If we distribute and simplify, we have area as a function of x alone:

A(x) = 4x^1/2 − x^3/2.

We can take the derivative with the power rule:

Aʹ(x) = 4(1/2)x^−1/2 − (3/2)x^1/2

The third (and final) step of Problem 16 said, “Find the maximum value of the area function A(x). What are the dimensions of the largest possible rectangular plot of grass? (Include units in your answers.)”

When we set Aʹ(x) equal to 0 to find critical values, we have to do a little bit of algebra. If, for example, we multiply both sides of the equation by 2x^1/2 to clear the fractions and simplify the exponents, we get the following equation to solve:

4 − 3x = 0

(You might want to check this to make sure you see how we get it.) It quickly follows that x = 4/3. That means that

Length = 4 − x = 4 − 4/3 = 8/3

and

Width = x^1/2 = (4/3)^1/2 = 2/(3^1/2)

The dimensions are in yards. If we multiply these together, we get an area of approximately 3.08 square yards.

And we are done.