Science:Math Exam Resources/Courses/MATH110/April 2017/Question 09/Solution 1

We continue from the hint. Let y be the height of the rectangle in the picture. We would like to solve for y in terms of x.

The small triangle on the bottom right is similar to the big triangle with dimensions 3 meters and 4 meters. The height of this triangle is y, and the width of this triangle is 3-x. We therefore have the following similarity equation:

${\frac {3}{3-x}}={\frac {4}{y}}.$

We multiply both sides by y and by 3-x:

$3y=4(3-x)$

And now, we divide by 3:

$y={\frac {4(3-x)}{3}}$

Because the rectangle has width x and height y, we have that the area of the rectangle is

$x{\frac {4(3-x)}{3}}$

We multiply the x into the numerator:

${\frac {4x(3-x)}{3}}$

and multiply the 4x into the (3-x) term:

${\frac {12x-4x^{2}}{3}}$

and rewrite this as the difference of two fractions:

$4x-{\frac {4}{3}}x^{2}.$

So we have that the area, A, is

$A=4x-{\frac {4}{3}}x^{2}.$

Because we want to maximize the area, we should take the derivative of A and set it equal to zero:

${\begin{aligned}{\frac {dA}{dx}}&=4-{\frac {8}{3}}x\\0&=4-{\frac {8}{3}}x\\{\frac {8}{3}}x&=4\\x&={\frac {12}{8}}\\x&={\frac {3}{2}}\end{aligned}}$

It remains to compute the y-value for each x. We computed that

$y={\frac {4(3-x)}{3}}$

If we plug in x = 3/2, we get

${\begin{aligned}y&={\frac {4(3-3/2)}{3}}\\y&={\frac {4(3/2)}{3}}\\y&={\frac {6}{3}}\\y&=2.\end{aligned}}$

This gives the dimensions

$x=3/2;y=2$

We should also check the endpoints of our domain. The smallest x can possibly be is zero, in which case the area is obviously zero. Similarly, the largest x can be is 3, in which case the area is also zero. Therefore, the point

$x=3/2;y=2$

is a genuine maximum.

Answer: $\color {blue}x=3/2;y=2$