## A fence is to be built to enclose a rectangular area of 450 square feet. The fence along three sides is to be made of material that costs ​

Question

A fence is to be built to enclose a rectangular area of 450 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.

The short side is ____ft and the long side is ______ ft.

in progress 0
2 weeks 2021-09-10T14:58:02+00:00 1 Answer 0

The short side is _15___ft and the long side is ___30___ ft.

Step-by-step explanation:

As fence is built in a rectangular area, so we can consider

Let

x be the length of the rectangle

y be width of the rectangle

Given area of rectangle is = 450 ft²

Formula for area of rectangle = Length x width

450 ft² =  xy

Solve for y

y = 450/x

now according to given condition

three sides of the fence costs \$5 per foot and for the fourth side costs\$15 per foot.

We have two condition either the fourth side be x or y

So condition 1:  Three sides =(x,y,x)       4-th side = y.

So  we can write as 5x,5y,5x  and  15 y

Cost C = 5x +5y+5x+15y

= 10x+ 5y+15y

= 5(2x+y) +15y—————-equation 1

= 10x +20y

= 10x + 20(450/x)

= 10x + 9000/x

For minimum cost, we can consider the cost to be 0

0 = 10x + 9000/x

Dividing and multiplying by -x/x

0 = -10 +9000/x²

10 = 9000/x²

10x² = 9000/

x²= 900

x = 30

so y = 450/x = 450/30= 15 ft

so adding the values of x and y in equation 1 we will have

cost C= 5(2x+y) +15y—————-equation 1

cost is          = 5(2(30)+15) +15(15)

=  \$600 is the cost

X= 30 y =15

So condition 2:  Three sides =(y,x,y)       4-th side = x.

So  we can write as 5y,5x,5y  and  15 x

Cost C = 5y +5x+5y+15x

= 5x+ 10y+15x

= 5(x+2y) +15x—————-equation 2

= 20x +10y

= 20x + 10(450/x)

= 20x + 4500/x

For minimum cost, we can consider the cost to be 0

0 = 20x + 4500/x

Dividing and multiplying by -x/x

0 = -20 +4500/x²

20 = 4500/x²

20x² = 4500

x²=  4500/20= 225

x = 15

so y = 450/x = 450/15= 30 ft

so adding the values of x and y in equation 2 we will have

cost C= = 5(x+2y) +15x—————-equation 2

cost is          = 5(15+2(30) +15(15)

=  \$600 is the cost

y= 30 x=15

so from both conditions satisfy the cost and the two sides are known as length and width

so dimension will be 15 ft by 30 ft