I don’t understand this question about finding the dimensions of a fenced piece of land. What are the steps to do this?

Question

A rectangular piece of land is to be fenced in using two kinds of fencing. Two opposite sides will be fenced using standard fencing that costs $6/m, while the other two sides will require heavy-duty fencing that costs $9/m. What are the dimensions of the rectangular lot of greatest area that can be fenced in for a cost of $9000?

answer: largest area is 93,750 m^2 when the width is 250m by 375m

thank you!

Answer 1

A = LW

2(9L+6W) = 9000

9L + 6W = 4500

3L+2W= 1500

2W = 1500-3L

W = 750-3L/2

A=LW = L(750-3L/2) = 750L -3L^2/2

A' = 750 - 3L = 0

3L = 750

L = 750/3 = 250 meters long

W = 750-3L/2 = 750-(3/2)250 = 375 meters wide

max A = 250(375) = 93,750 square meters

dimensions are 250 meters by 375 meters to have maximum area for $9,000

Answer 2

Call x the cheap fencing and y the expensive fencing.

2x*6 + 2y*9 = 9000

12x + 18y = 9000 (divide all terms by 6)

2x + 3y = 1500

3y = 1500 - 2x

y = (1500 - 2x)/3

A = xy = x(1500-2x)/3 = (1/3)* 1500x - 2x^2 = 500x - (2/3)x^2 (This is an upside down parabola, so the vertex is the max). Differentiate or use -b/2a.

a = -2/3 b = 500

-b/2a = 500/4/3 = 500*3/4 = 1500/4 = 375

y = (1500 - 2(375)/3 = 750/3 = 250

The max size is 375 by 250

If you want to use derivatives, do this.

A = 500x - (2/3)x^2

A' = 500 -4x/3

A'' = -4/3

A' = 0 = critical point: 500 - 4x/3 = 0

-4x/3 = -500

x = -500*(-3/4) = 375

A''(x=375) = -4/3 which is < 0, so it's a max.

y = (1500 - 2(375)/3 = 750/3 = 250

The max size is 375 by 250