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olga_2 [115]
3 years ago
14

A rectangular field with one side along a river is to be fenced. Suppose that no fence is needed along the river, the fence on t

he side opposite the river costs $20 per foot, and the fence on the other sides costs $5 per foot. If the field must contain 80,000 square feet, what dimensions will minimize costs?
a) Side Parallel to the River___________ft
b) Each of the other sides _____________ft
Mathematics
1 answer:
Andreas93 [3]3 years ago
4 0

Answer:

a) Side Parallel to the river: 200 ft

b) Each of the other sides: 400 ft

Step-by-step explanation:

Let L represent side parallel to the river and W represent width of fence.

The required fencing (F) would be F=L+2W.

We have been given that field must contain 80,000 square feet. This means area of field must be equal to 80,000.

LW=80,000...(1)

We are told that the fence on the side opposite the river costs $20 per foot, and the fence on the other sides costs $5 per foot, so total cost (C) of fencing would be C=20L+5(2W)\Rightarrow 20L+10W.

From equation (1), we will get:

L=\frac{80,000}{W}

Upon substituting this value in cost equation, we will get:

C=20(\frac{80,000}{W})+10W

C=\frac{1600,000}{W}+10W

C=1600,000W^{-1}+10W

To minimize the cost, we need to find critical points of the the derivative of cost function as:

C'=-1600,000W^{-2}+10

-1600,000W^{-2}+10=0

-1600,000W^{-2}=-10

-\frac{1600,000}{W^2}=-10

-10W^2=-1,600,000

W^2=160,000

\sqrt{W^2}=\pm\sqrt{160,000}  

W=\pm 400

Since width cannot be negative, therefore, the width of the fencing would be 400 feet.

Now, we will find the 2nd derivative as:

C''=-2(-1600,000)W^{-3}

C''=3200,000W^{-3}

C''=\frac{3200,000}{W^3}

Now, we will substitute W=400 in 2nd derivative as:

C''(400)=\frac{3200,000}{400^3}=\frac{3200,000}{64000000}=0.05

Since 2nd derivative is positive at W=400, therefore, width of 400 ft of the fencing will minimize the cost.

Upon substituting W=400 in L=\frac{80,000}{W}, we will get:

L=\frac{80,000}{400}\\\\L=200

Therefore, the side parallel to the river will be 200 feet.

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