What values of x and y would result in the maximum area that you can enclose?

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? x has three sides and y has four with two of them having a +5. the image is two rectangles adjacent to each other. 410=3x+4y+10

X

_____________

| | Y+5

|____________ |

|____________ | Y

Answer 1

Yes, Thank you Lawrence!

As you said, the perimeter is 410 = 3x+4y+10

The area function is (2y+5)*x = Area

Solving the perimeter for y:

410 - 3x - 10 = 4y

(410-3x-10)/4 = y

So the area function as a function of x is:

Area = A(x) = [2(410-3x-10)/4 + 5]*x

= [(410-3x-10)/2 + 5]*x

= [205 - (3/2)x - 5 + 5]*x

= (205 - (3/2)x)*x

= 205x - (3/2)x^2

Maximizing:

0 = dA/dx = 205 - 3x

3x = 205

x = 205/3

Then y = (410 - 3x -10)/4 = (410 -205 -10)/4 = 195/4

The max area is (2*195/4 + 5) (205/3) = (195/2+5)(205/3)

= (195/2 + 10/2)(205/3)

= (205/2)(205/3) = 7004 and 1/6

Answer 2

maximize (y)(x)+(y+5)x = (2y+5)x subject to 3x + 4y + 10 = 410, or 3x + 4y = 400

We can either use lagrange multipliers or, as 3x + 4y = 400, y = 100 - 3/4x

maximize A = 2x(100 - 3/4x ) + 5x = 205x - 3/2x²

dA/dx = 205 - 3x

x = 205/3

y = 100 - 3/4(205/3) = 48 3/4