Question

You have land that you would like to use to create two distinct fenced-in areas in the shape given below. You have 410 meters of fencing materials to use. What values of x and y would result in the maximum area that you can enclose

Answer 1

Answer:

$x = \frac{205}{3}$

$y =\frac{195}{4}$

Step-by-step explanation:

Given

$p = 410$ --- perimeter

See attachment for fence

Required

x and y

The perimeter of the fence is:

$p = 2(x + y +5 + y) +x$

Open bracket

$p = 2x + 2y +10 + 2y +x$

Collect like terms

$p = 2x+x + 2y + 2y+10$

$p = 3x + 4y+10$

Substitute: $p = 410$

$3x + 4y+10 =410$

Make 4y the subject

$4y =410-10-3x$

$4y =400-3x$

Make y the subject

$y =\frac{400-3x}{4}$

The area (A) of the fence is:

$A = (y + y + 5) * x$

$A = (2y + 5) * x$

Substitute: $y =\frac{400-3x}{4}$

$A = (2*\frac{400-3x}{4} + 5) * x$

$A = (\frac{400-3x}{2} + 5) * x$

Take LCM

$A = (\frac{400-3x+10}{2}) * x$

Solve like terms

$A = (\frac{410-3x}{2}) * x$

Open bracket

$A = \frac{410x-3x^2}{2}$

Remove fraction

$A = 205x-1.5x^2$

Differentiate both sides

$A' = 205 - 3x$

To maximize; set $A' =0$

$205 - 3x =0$

Solve for 3x

$3x = 205$

Solve for x

$x = \frac{205}{3}$

Recall that: $y =\frac{400-3x}{4}$

So, we have:

$y =\frac{400-3*205/3}{4}$

$y =\frac{400-205}{4}$

$y =\frac{195}{4}$