1. ## Problem with Optimization

This question really threw me for a loop:
A closed rectangular container with a square base is to have a volume of 2250 cubic inches. The material for the top and bottom of the container will cost $2 per square inch and the material for the sides will cost$3 per square inch. Find the dimensions of the container of least cost.

What I have so far:
$\displaystyle V=x^2h = 2250$ so $\displaystyle h = \frac{2250}{x^2}$

$\displaystyle S = 2x^2 + 4xh$ so $\displaystyle S = 2x^2 + 4x * \frac{2250}{x^2} = 2x + \frac{9000}{x}$

But when I thought about trying to find the feasible domain, I got stumped. The question doesn't specify a minimum or maximum value for either a side of the base or the height of the container. So, theoretically, x would get infinitely large as h got infinitely small and vice versa. Is there a generally accepted thing to do in this situation or am I doing something wrong here?
Thanks for the help

Oh, and i had a few more equations relative to the problem:
I separated the equation for surface area into $\displaystyle S = s_{1} + s_{2}$ where S is the total surface area,$\displaystyle s_{1}$ is the area of the base($\displaystyle 2x^2$), and $\displaystyle s_{2}$ is the area of a side($\displaystyle 4xh$).
I then made the equation $\displaystyle P = 2s_{1} + 3s_{2}$ where P is the total price. But I didn't know what to do with all this...

Edit: I suppose I'll just attempt to finish the problem with a feasible domain of 0 < x < $\displaystyle \infty$, although I won't be able to test for the endpoints... oh well.

2. You were close, but the question asks for price minimisation, not surface area minimisation.

$\displaystyle x^2h = 2250$

$\displaystyle \text{Price} = 2x^2 \cdot \$2 + 4xh \cdot \$3$

Make the substitution:

$\displaystyle \text{Price} = 2x^2 \cdot \$2 + 4x(\frac{2250}{x^2}) \cdot \$3$

$\displaystyle \text{Price} = 4x^2 + \frac{27000}{x}$

Now differentiate:

$\displaystyle P(x) = 4x^2 + \frac{27000}{x}$

$\displaystyle P'(x) = 8x - \frac{27000}{x^2}$

Set P'(x) to 0 and solve for x.

3. Ah, yes. It makes so much more sense now. Thanks for the help.
In the end, i got x = 15 and h = 10. So a box with dimensions of 15" x 15" x 10" would do the trick.

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### a closed rectangular box having a square base is to be constructed so that it's volume is 2000 in³. if the material for the top and bottom of the box costs 20 pesos per square inch while the material for the sides costs 10 pesos per square inch, what sho

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