A box with a square base has a volume of 1500 cubic inches...

A box with a square base has a volume of 1500 cubic inches. The material for the top and bottom of the box costs 1.5 cents/square inch and the material for the sides of the box costs 2 cents/square inch. Let x denote the length of a side of the square base and let h denote the height of the box. * A. Define a function for the cost of the material for the box in terms of x * B. What is the domain of the function in part A? * C. Using calculus, find the dimensions of the box that will minimize the cost of the material used. Be sure to show your answer does give a minimum.

Transcript

00:02 All right, so we need to, first of all, set up a cost function.

00:06 So the cost is going to be two tops times 1 .5 times the area of the top, and the area of the top is x squared.

00:18 That's for the top and the bottom.

00:20 Then we're going to add to that the cost for the sides.

00:22 There are four sides.

00:24 There are two cents per square inch, and the area there is xh.

00:28 Okay.

00:30 The domain for these, since everything's multiplied together to work this, out there's no limit to how long they can be except for they can't make a zero right so you have to have x has to be greater than zero and h has to be greater than zero they can be infinitely large it's just not realistic okay but you can't do that and then see to find the the minimum values of what we're looking for here we're looking for the dimensions that will make the minimum cost we're going to show the volume first the volume is x squared h and that is a thousand or fifteen hundred rather okay, so that means that h can be solved as 1500 over x squared.

01:16 We're now going to take the cost equation and replace the h with the 1500 or x squared.

01:22 So we get 3x squared plus 8x times 1500 over x squared.

01:33 So the cost will equal 3x squared.

01:37 I'm going to actually factor out the 3 to make it x squared plus.

01:41 If i take a 3 out of here, that leaves 8 times 500, so it's 4 ,000 over x squared.

01:47 That's your cost.

01:49 We're then going to take a derivative of that to get the maximum minimum.

01:54 So that's going to be 3, and then it'll be 2x minus 400 or 4 ,000 over x squared.

02:09 I'm sorry, i did that wrong.

02:10 This is x here.

02:12 The x is will cancel.

02:13 It's just x.

02:13 That's x squared, which is now i'm going to take the two out to give me 6...

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