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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?

Respuesta :

Answer:

[tex] C'(y) = 50 -\frac{3900000}{y^2}=0 [/tex]

And we can solve for y and we got:

[tex] y = \sqrt{\frac{3900000}{50}}= 279.285[/tex]

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

[tex] x= \frac{60000}{279.285}= 214.834[/tex]

So then the minimum cost for this case would be:

[tex] C = 50*279.285 + 65*214.834 = 27928.49[/tex]

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:

[tex] xy = 60000[/tex]

If we solve for x we got:

[tex] x = \frac{60000}{y}[/tex]  (1)

Now we can define the cost function like this:

[tex] C = 2*(25)*y + 25 x +40 x[/tex]

[tex] C(x,y) = 50 y + 65 x[/tex]

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

[tex] C(y) = 50 y + 65(\frac{60000}{y})[/tex]

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

[tex] C'(y) = 50 -\frac{3900000}{y^2}=0 [/tex]

And we can solve for y and we got:

[tex] y = \sqrt{\frac{3900000}{50}}= 279.285[/tex]

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

[tex] x= \frac{60000}{279.285}= 214.834[/tex]

So then the minimum cost for this case would be:

[tex] C = 50*279.285 + 65*214.834 = 27928.49[/tex]

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