algebra 2
posted by Rachel .
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?

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(602x)
= 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((x15)^2  225)
= 2(x15)^2 + 450
so the area is a maximum when x = 15
and then y = 6030 = 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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