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.

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

Answers ( )

    0
    2021-09-10T14:59:43+00:00

    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  

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