Calculus problem

Question

A) Find the dimensions of the rectangle with area 361 square inches that has minimum perimeter, and then find the minimum perimeter.

L = , W = , Minimum Perimeter = .

B) Find the dimensions of the rectangle with perimeter 92 inches that has maximum area, and then find the maximum area.

L = , W = , Maximum area = .

Answer 1

A)

Area, A=LW=361 in²

W=361/L

Perimeter, P=2L+2W=2L+722/L

dP/dL = 2-722/L²

Set dP/dL to zero and solve for L

2-722/L²=0

2L²=722 (L≠0)

L=√361=19

W=361/19=19

P=2(19)+2(19)

B)

P=2L+2W=92

W=(92-2L)/2 = 46-L

A = LW = L(46-L)

dA/dL = 46-2L

L=23

W=46-23

A=LW = 23²