Afence 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 and the long side is ft.

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) +15yequation 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) +15yequation 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) +15xequation 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) +15xequation 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