ALWAYS START OPTIMIZATION PROBLEMS BY DRAWING DIAGRAMS! see below, i drew some for you.

2x + 2y = 16 = P ..........P is for perimeter1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

we want Area = A = xy to be a maximum

since 2x + 2y = 16

=> x = 8 - y

so A = (8 - y)y = 8y - y^2

for max, we want A' = 0

=> A' = 8 - 2y = 0

=>y = 4

But x = 8 - y

=>x = 4

this is very similar to the last one2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

x + 2y = 40

want A = xy to be max

since x + 2y = 40

=> x = 40 - 2y

=> A = (40 - 2y)y = 40y - 2y^2

for max, A' = 0

=> A' = 40 - 4y = 0

=>y = 10

but x = 40 - 2y

=>x = 20

A = xy = 323) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?

we want Perimeter = P = 2x + 2y to be minimum

since xy=32

=> x = 32/y

=> 2(32/y) + 2y = P

=> P = 64/y + 2y = 64y^-1 + 2y

for min, P' = 0

=> P' = -64y^-2 + 2 = 0

=> -64 + 2y^2 = 0

=> 2y^2 = 64

=> y^2 = 32

=> y = +/- sqrt(32)

soy = sqrt(32)since we can't have negative area

but x = 32/y

=> x = 32/sqrt(32)

=>x = sqrt(32)