A rectangular parking area measuring 6000 ft squared is to be enclosed on three sides using chain-link fencing that costs $5.50 per foot. The fourth side will be a wooden fence that costs $6 per foot. What dimensions will minimize the total cost to enclose this area, and what is the minimum cost?

Question

Karina Frazier

Mathematics

31 March, 05:40

A rectangular parking area measuring 6000 ft squared is to be enclosed on three sides using chain-link fencing that costs $5.50 per foot. The fourth side will be a wooden fence that costs $6 per foot. What dimensions will minimize the total cost to enclose this area, and what is the minimum cost?

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Answers (1)

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Maggie · Answer 1

length x = 79.2 ft

width y = 75.75 ft

minimum cost P = Rs 1742.41212

Step-by-step explanation:

Let the length of the parking area be 'x', and the width be 'y'.

Then, we can write the following equations:

Then, Area of the park: A = x*y = 6000

Now, Price of the fences P = 2*5.5x + 5.5y + 6y

P = 11x + 11.5y

From the first equation, we have that y = 6000/x

Using this value in the equation for P, we have:

P = 11x + 11.5*6000/x = 11x + 69000/x

To find the minimum of this function, we need to take its derivative and then make it equal to zero:

⇒ dP/dx = 11 - 69000/x^2 = 0

⇒x^2 = 69000/11

x = 79.20 ft

This is the x value that gives the minimum cost.

Now, finding y and P, we have:

x*y = 6000

y = 6000/79.2 = 75.75 ft

P = 11x + 69000/x = Rs 1742.41212