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.

Answer 1

Answer:

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

Adding value of y 450/x

        = 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

Adding value of y= 450/x

        = 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