Question

A fence is to be built to enclose a rectangular area of 800 square feet. The fence along three sides is to be made of material that costs ​$5 per foot. The material for the fourth side costs ​$15 per foot. Find the dimensions of the rectangle that will allow for the most economical fence to be built.

Answer 1

Answer:

dc/dx = 0 when x = 20. hence fence is 20×40

Step-by-step explanation:

Given data:

Rectangular Area = 800 ft^2

three side material cost = $5 per ft

fourth side material cost is $15 per ft

let expensive side is x and other dimension assume to be y

Then cost is c

c = 5(x+ 2y) + 15 x

xy = 800, so y = 800/x

c =5(x+ 1600/x^2) + 15x

for minimum when dc/dx = 0

$\frac{dc}{dx} = 5(1-\frac{1600}{x^2}) + 15 = 0$

$5(1-\frac{1600}{x^2}) + 15 = 0$

$(1-\frac{1600}{x^2}) + 3 = 0$

$(1-\frac{1600}{x^2}) = - 3$

$x^2 - 1600 = -3x^2$

$4x^2 - 1600 = 0$

(2x-40)(2x+40) = 0

x = 20

dc/dx = 0 when x = 20. hence fence is 20×40