algebra 2

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A dog kennel is to be constructed alongside a house using 60 m of fencing. Find the dimensions that will yield the maximum area.

How do you solve this problem?

• algebra 2 -

I assume you are implying that we don't need a fence along the house.

let the length be y m (the single side)
let the width be x m

so we know that 2x + y = 60
or y = 60 - 2x

Area = xy
= x(60-2x)
= -2x^2 + 60x

At this point I don't if you know Calculus or not.
If you do, then
d(Area)/dx = -4x + 60
= 0 for a max of area
-4x + 60 = 0
x = 15
then y = 30

if you don't take Calculus, we have to complete the square
area = -2(x^2 - 30x)
= -2(x^2 - 30x + 225 - 225)
= -2((x-15)^2 - 225)
= -2(x-15)^2 + 450

so the area is a maximum when x = 15
and then y = 60-30 = 30

by either method,
the width has to be 15 m, and the length has to be 30 m for a maximum area of 15(30) or 450 m^2

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