Question

A rectangular garden is bounded on one side by a house and is to be
fenced on the other three sides. Fencing material costs $16 per foot for the
side parallel to the house and $20 per foot for the other two sides. What
are the dimensions of the garden of largest possible area if $1600 is to be
spent for fencing material?

Answer 1

Let the side parallel to the house is x. The other side is y.
Total perimeter to be fenced is the three sides = x + y + y = x +2y
Cost of fencing side which is y foot is 20*y
Cost of fencing side parallel to house( x foot) = 16*x
Total cost is 2* (20y) + 16x
Total cost = 40y +16x.
Cost = 1600
1600 = 40y +16x -------------(i)
Divide the equation (i) by 4
400 = 10y + 4x
(400 - 10y)/4 = x ------------(ii)

Area = x*y.
We want to maximize area
Plug in value of x from (ii)
(400-10y)/4 * y = area
100y - 5/2 y^2 = area
Differentiating with respect to y, we get 100 - 5y =A'(y)
100 -5y = 0 , solving we get y = 20.
x = (400 -10y)/4 = (400 -200)/4 = 200/4 = 50
Dimensions are 50 ft and 100 ft

Answer 2

Rita you made a mistake in your final answer. The math is there though. Consider the $1600 budget:
\[2(20*50)+16(100)=3600\]
If you plug the 20 in you solved for y:
\[1600=40(20)+16x\]
\[1600=800+16x\]
\[800=16x\]
\[x=50\]
So two 20 ft runs and a single 50 ft run to get the maximum area. Like I said you did all the hard lifting :)