Question

A 60,000 square meter rectangular yard is to be enclosed on three sides by wood fencing that costs $25.00 per meter and on the fourth side with a stone wall that will cost $40.00 per meter. What is the minimum cost of enclosing this yard as described?

Answer 1

Answer:

$C'(y) = 50 -\frac{3900000}{y^2}=0$

And we can solve for y and we got:

$y = \sqrt{\frac{3900000}{50}}= 279.285$

And using condition (1) we can solve for x and we got:

$x= \frac{60000}{279.285}= 214.834$

So then the minimum cost for this case would be:

$C = 50*279.285 + 65*214.834 = 27928.49$

Explanation:

For this case the graph attached illustrate the problem for this case

We know that the total area is 60000, so then we have:

$xy = 60000$

If we solve for x we got:

$x = \frac{60000}{y}$  (1)

Now we can define the cost function like this:

$C = 2*(25)*y + 25 x +40 x$

$C(x,y) = 50 y + 65 x$

We can use the condition (1) and if we replace in the cost function we have:

$C(y) = 50 y + 65(\frac{60000}{y})$

Since we need to minimize the cost, we can derivate the function in terms of y and we got:

$C'(y) = 50 -\frac{3900000}{y^2}=0$

And we can solve for y and we got:

$y = \sqrt{\frac{3900000}{50}}= 279.285$

And using condition (1) we can solve for x and we got:

$x= \frac{60000}{279.285}= 214.834$

So then the minimum cost for this case would be:

$C = 50*279.285 + 65*214.834 = 27928.49$