Respuesta :
Answer:
[tex]18.97 \text{ and }12.65\text{ft}[/tex]
Step-by-step explanation:
GIVEN: A fence is to be built to enclose a rectangular area of [tex]240[/tex] square feet. The fence along three sides is to be made of material that costs [tex]6[/tex] dollars per foot, and the material for the fourth side costs [tex]12[/tex] dollars per foot.
TO FIND: Find the dimensions of the enclosure that is most economical to construct.
SOLUTION:
Area of rectangular fence [tex]=240\text{ sq. ft}[/tex]
let the length of fence [tex]=x[/tex]
let the width of fence [tex]=y[/tex]
let [tex]x[/tex] be the smaller side
Area of rectangular fence enclosure [tex]=xy[/tex]
[tex]xy=240[/tex]
[tex]y=\frac{240}{x}[/tex]
cost of fence along three sides [tex]=6 \text{ dollars per feet}[/tex]
cost of fence along fourth side [tex]=12 \text{ dollars per feet}[/tex]
length of fence [tex]=2x+2y[/tex]
cost of fence building [tex]=6(2x+y)+12y[/tex]
[tex]=12x+18y[/tex]
putting value of [tex]y[/tex]
[tex]=12x+18\times\frac{240}{x}[/tex]
[tex]=12x+4320x^{-1}[/tex]
to find minimum value differentiating the equation
[tex]12-\frac{4320}{x^{2} }=0[/tex]
[tex]x^{2} =\frac{4320}{12}[/tex]
[tex]x^{2} =360[/tex]
[tex]x=18.97\text{ft}[/tex]
[tex]y=12.65\text{ft}[/tex]
Hence the dimensions of the enclosure that is most economical to construct are [tex]18.97\text{ft}[/tex] and [tex]12.65\text{ft}[/tex]
The most economical to construct would have a length of 19 ft. and width of 12.6 ft.
Let x represent the length of the rectangle and y represent the width of the rectangle.
Since the area is 240 square feet, hence:
xy = 240
y = 240/x
The total cost (C) of material is:
C = 6x + 6y + 6x + 12y
C = 12x + 18y
C = 12x + 18(240/x)
C = 12x + 4320/x
The minimum cost is at dC/dx = 0, hence:
dC/dx = 12 - 4320/x²
12 - 4320/x² = 0
12 = 4320/x²
12x² = 4320
x² = 360
x = 19 ft
y = 240/x = 240/19 = 12.6 ft
The most economical to construct would have a length of 19 ft. and width of 12.6 ft.
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