Respuesta :
Answer with explanation:
Let the length of rectangular fence = x meters
Let the width of the rectangular fence = y meters
Thus
[tex]Area=x\times y=210ft^{2}...........(i)[/tex]
Now Let the total cost to construct the fence be 'C' can be obtained if we construct the shorter side with material with cost $12 per foot
[tex]\therefore C=5\times (2y+x)+12\times x\\\\C=10y+17x...............(ii)[/tex]
Using value of x from the equation i into equation ii we get
[tex]C=10y+17\times \frac{210}{y}[/tex]
hence to minimize the cost we differentiate both sides with respect to 'y' and equate the result to zero thus we get
[tex]C=10y+17\times \frac{210}{y}\\\\\frac{dC}{dy}=\frac{d}{dy}(10y+\frac{3570}{y})\\\\0=10-\frac{3570}{y^{2}}\\\\\therefore y=\sqrt{\frac{3570}{10}}=18.894feet[/tex]
Thus [tex]x=\frac{210}{18.894}=11.11feet[/tex]
Thus value of length = 11.114 feet and value of width = 18.894 feet.
Answer:
Length of enclosure =18.89 foot
Width of enclosure=11.12 foot
Step-by-step explanation:
We are given that a fence is to be built to enclose a rectangular area 210 square feet.
Fence along three sides is to be made of material that costs 5 dollars per foot
and the material for the fourth side costs 12 dollars per foot.
We have to find the dimension of the enclosure that is most economical
Let x be the length and y be the width of enclosure
We know that area of rectangle=[tex]x\timesy[/tex]
[tex]xy=210[/tex]
[tex]y=\frac{210}{x}[/tex]
Cost of four sides =[tex]2(5x)+5(y)+12(y)[/tex]
Total cost=[tex]10x+17y[/tex]
C=[tex]10x+17\cdot\frac{210}{x}[/tex]
C=[tex]10x+\frac{3570}{x}[/tex]
Differentiate w.r.t x
[tex]\frac{dC}{dt}=10-\frac{3570}{x^2}[/tex]
Substitute[tex]\frac{dC}{dx}=0[/tex]
[tex]10-\frac{3570}{x^2}=0[/tex]
[tex]\frac{3570}{x^2}=10[/tex]
[tex]x^2=357[/tex]
[tex]x=\sqrt{357}[/tex]
x=18.89
Differentiate w.r.t x
[tex]\frac{d^2C}{dx^2}=\frac{7140}{x^3}[/tex]
Substitute x=18.89
Then we get
[tex]\frac{d^2C}{dx^2}=\frac{7140}{(18.89)^3} >0[/tex]
Hence, the cost is minimum.
Length of enclosure =18.89 foot
Width of enclosure=[tex]\frac{210}{18.89}=11.12 foot[/tex]
Hence, the dimension of the enclosure that is most economical to construct
Length=18.89 foot and width=11.12 foot
