Calculus. Help please!!!

Question

A homeowner wants to enclose a 2,900 square foot rectangular garden by a fence in his backyard. If 3 sides of the fence cost $4.20 per foot and the 4th side costs $5.80 per foot, find the dimensions that will minimize the cost of building the fence and the minimum cost of its construction

Answer 1

x is length, y is width. Area A(x,y) = xy = 2900 Cost is C(x,y) = 2x*4.20 +y*4.20 +y*5.80 = 8.40x + 10.00y.

y = 2900/x, C(x) = 8.40x +29000/x

Derivative C'(x) = 8.40 - 29000/x²

C'(x) = 0 to get critical value of x

8.40 - 29000/x²= 0, x³= 29000/8.40, x = (29000/8.40)^1/3 ≅ 15.11 ft.

This x give C(x) minimum because of second derivative Test f""(x) = 58000/x³ > 0 at x = 15.11.

Then y = 2900/15.11 = 191.28 ft

minimum cost C = 8.40* 15.11 + 10.00 * 191.28 ≅ $2039.72

Answer 2

A=s₁*s₂ so 2900 = (s₁)(s₂) or s₂ = 2900/s₁

Also s₁ = s₃ and s₂ = s₄

C = 4.20(s₁) + (4.20)(s₂) + 4.20(s₃) + 5.80(s₄)

C = 4.20(s₁ + s₂ + s₃) + 5.80(s₄)

C(s₁) = 4.20(s₁ + (2900/s₁) + s₁) + 5.80(2900/s₁)

C(s₁) = 8.4s₁ + 11890/s₁ + 16820/s₁

C(s₁) = 8.4s₁ + 28710/s₁

To minimize, take the first derivative and set it equal to zero

C'(s₁) = 8.4 - 28710/s₁²

8.4 - 28710/s₁² = 0

8.4 = 28710/s₁²

s₁² = 28710/8.4

s₁ = 58.46

s₂ = 49.40

So, the dimensions to minimize cost are 58.46 ft x 49.40 ft