Calculus Question pls help Im stuck with this problem. (problem in description)

Question

A landscape architect plans to enclose a 3000 square foot rectangular region in a botanical garden. She will use shrubs costing $15 per foot along three sides and fencing costing $20 per foot along the fourth side. Find the dimensions that minimize the total cost. The fenced side is___ feet. Make sure to prove that you've found the absolute minimum in calc terms. (Round the answer to the 1 decimal place.)

Answer 1

20L + 15L + 15(2W) = Cost = C

LW = 3000

W=3000/L

35L + 30(3,000/L) = C

C(L) = 35L + 90,000L^-1

take the derivative of C(L) and set equal to zero, solve for L

C' = 35 - 90000L^-2 = 0

35=90,000/L^2

L^2 = 90,000/35

L= 300/sqr35 = 300sqr35/35= about 50.71 feet of fencing one 1 side and of shrubs on the opposite side

W =3000/300/sqr35 = 10sqr35=59.16 feet of shrubs on 2 sides

W=59.2

L=50.7

WL = square feet

but rounding to one decimal gives WL=3001 square feet

at cost of 30(59.2) + 15(50.7) + 20(50.7)

= 1776 + 1774.5 = $3550.50

but 30(60)+35(50)=1800+1750= $3550.00, 50 cents less

and 60x50=3000 square feet

30(59.16)+35(50.71) = 1774.8 + 1774.85 = $3549.65 the minimum cost with 59.16 by 50.71 feet

rounding errors make a few cents difference

more exact dimensions are 59.16079783 by 50.70925528 feet for minimum cost

for calculus on the absolute minimum, take the 2nd derivative

or look at the end points

60x50 feet is virtually the cost minimizing dimensions.

look at some other simple numbers on either side and you'll find higher cost. It's

a local minimum, but the endpoints show it's also an absolute minimum

Answer 2

3L*15 + 20W = C (cost)

LW=3000 => W=3000/L

45L + 60000/L = C

dC/dL = 45 - 60000/L² = 0

L is about 36' ...but I will leave the arithmetic to you