Applied Maxima and Minima

[courtrigrad]

Hello all

You are planning to make an open box from a 30- by 42 inch piece of sheet metal by cutting congruent squares from the corners and folding up the sides. You want the box to have the largest possible volume.

(a) What size square should you cut from each corner? (gice side length of square)

(b) What is the largest possible volume the box will have

I know $$V(x) = x(30-2x)(42-2x)$$

So $$V'(x) = (30 - 2 x) (42 - 2 x) - 2 x (42 - 2 x) - 2 x (30 - 2 x)$$

I find the critical points, however how do I find the maximum volume. Also I am not sure how you would find what square size you should cut. Shouldn't you find the maximum volume?

Thanks

 

[MathStudent]

Both questions are really solved simultaneously. You are trying to find the value of x that makes V(x) take on the greatest value, thus that value for x will be the length of the side of the square you cut.

Here are some questions to get you started:
How do you find local extrema of a function?
How do you find absloute max?
what values of x make V(x) meaningless?

 

[wais]

for sharing this problem with us! Finding the maximum volume of the box is indeed the goal of this problem. To find the maximum volume, you need to use the first derivative test to identify the critical points, and then use the second derivative test to determine whether the critical points are maximum or minimum points.

To find the size of the square that should be cut from each corner, you need to set the first derivative equal to 0 and solve for x. This will give you the value of x that maximizes the volume.

If you are having trouble finding the critical points or determining the maximum volume, I suggest reviewing the process of finding extrema using derivatives and also checking your calculations. This problem can also be solved using graphing software or a graphing calculator to visualize the function and its extrema.

I hope this helps and good luck with your problem-solving!