Optimization problem (Calculus)

Question

A rectangular pigpen will be surrounded by a fence and then divided into two sections by a block wall. The area of the pigpen must be 50 m^2 . Fencing costs $20/m. The block wall costs $35/m. What should the dimensions be to minimize costs? Domain: x>0

Answer 1

Let x: length of the pigpen

Let y: width of the pigpen (the block wall will also be y m wide, to run parallel to these 2 sides)

Objective function (Quantity to be optimized): Cost = 40x + 40y + 35y = 40x + 75y

Constraint: A = xy = 50 : . y = 50/x. We substitute this expression in for y in the cost function to get cost as a function of x only:

C(x) = 40x + 3750 / x Now, we differentiate w respect to x and set the derivative = 0 and solve:

C^'(x) = 40 - 3750x^-2 = 0

40x² - 3750 = 0

x² = 375/4

x = √375 / 2 = 5√15 / 2

y = 100 / √375 = 20 / √15

Answer 2

Area = 50 m^2 = Lw = Length times width

L=50/w

width = block wall dimension

there's 4 widths, 2 of the fence, 2 of the block wall

2 lengths of the fence

fence costs $20 per meter

wall costs $35 per meter

2w(35) +2(20)w) + 2L(20) = Cost

110w + 40(50/w) = C(w)

110w+2000/w= C(w)

take the derivative of the Cost function and set = zero

C'(w) = 110 -2000/w^2 = 0

110w^2 -2000 = 0

11w^2 -200 =0

w^2 = 200/11

w = sqr(200/11) = 10sqr(2/11)= about 4.26 meters wide

L= 50/w = 50/(sqr200/11)=50/10sqr(2/11) = 5/sqr(2/11) = about 11.73 meters Long

4.26 x 11.743 = about 50 m^2

dimensions are about 4 1/4 by 11 3/4 meters