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

`x = (205)/(3)`

`y =(195)/(4)`

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 =(400-3x)/(4)`

The area (A) of the fence is:

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

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

Substitute:
`y =(400-3x)/(4)`

`A = (2*(400-3x)/(4) + 5) * x`

`A = ((400-3x)/(2) + 5) * x`

Take LCM

`A = ((400-3x+10)/(2)) * x`

Solve like terms

`A = ((410-3x)/(2)) * x`

Open bracket

`A = (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 = (205)/(3)`

Recall that:
`y =(400-3x)/(4)`

So, we have:

`y =(400-3*205/3)/(4)`

`y =(400-205)/(4)`

`y =(195)/(4)`