Question

(1 point) A fence is to be built to enclose a rectangular area of 240 square feet. The fence along three sides is to be made of material that costs 5 dollars per foot, and the material for the fourth side costs 15 dollars per foot. Find the dimensions of the enclosure that is most economical to construct. Dimensions: x

Answer 1

Answer:

Length = 19 ft

Width = 12.63 ft

Step-by-step explanation:

Let the length be 'L' and width be 'w'.

Given:

Area of the rectangular fence (A) = 240 ft²

Fencing cost for three sides = $5 per foot

Fencing cost for fourth side = $15 per foot

Area of the fence is given by the formula:

$A=Lw\\\\w=\frac{A}{L}\\\\w=\frac{240}{L}----(1)$

Now, Perimeter of first three sides = $L + L + w = 2L + w$

Cost of fencing three sides = Cost per foot × Perimeter of 3 sides

$C_1=5(2L+w)=10L+5w$

Cost of fencing fourth side = Cost per foot × length of fourth side

$C_2=15w$

Total cost of fencing, $C=C_1+C_2=10L+5w+15w=10L+20w$

Now, replace 'w' from equation (1). This gives,

$C=10L+\frac{15\times 240}{L}\\\\C=10L+\frac{3600}{L}$

Now, to minimize the total cost, the derivative of total cost with any of the dimensions must be 0.

Differentiating both sides with respect to length 'L', we get:

$\frac{dC}{dL}=\frac{d}{dL}(10L+\frac{3600}{L})=0\\\\10-\frac{3600}{L^2}=0\\\\\frac{3600}{L^2}=10\\\\L^2=\frac{3600}{10}\\\\L=\sqrt{360}=18.97\approx 19\ feet$

Therefore, the length is 19 feet.

Now, the width is determined using equation (1). So,

$w=\frac{240}{19}=12.63\ ft$

Therefore, the dimensions are length equal to 19 ft and width equal to 12.63 ft.