Folded boxes

a. Squares with sides of length ...

Question

Folded boxes

a. Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 5 ft by 8 ft. The resulting piece of cardboard is then folded into a box without a lid. Find the volume of the largest box that can be formed in this way.

Accepted Answer

Welcome back, everyone. In this problem, a square piece of cardboard with a side length of 10 ft as squares with sides of length X cut out from each corner. The remaining flaps are folded up to form an open top box. Find the value of X that minimizes the volume of the box and determine this maximum volume. Now, if we're going to find first the value of X that maximizes the volume and subsequently the maximum volume, then we need to understand exactly what's going on with our square piece of cardboard here. So let's do a quick sketch. So let's say that this is our square piece of cardboard, OK. And for our cardboard notice our problem tells us it has squares with sizes of length X that are cut out from each corner. So in other words, We can put our smaller squares here and we know that they have side lengths of X, OK? So it's not gonna be exact in this sketch, but essentially this is what we expect our cardboard to look like. And now here in that case then. That means that for this square, it's length L is going to be equal to, let me, I like that this length here. And Is gonna be, we know it has a side length of 10 ft, so the length is gonna be equal to 10 minus 2X and because it's a square, that's going to be the same for the width. It's gonna be 10 minus 2X, OK? And we know then that the height. E of our box. OK, let me just clean this up right here. We know that the height of our box, so let's write that down. For our box. OK. The length equals the width, which is 10 minus 2 X, and our height H is going to be equal to X. Know that we have this information. Then how can we find the volume of the box? We'll recall that the volume, OK, is gonna be equal to the length multiplied by the width multiplied by the height. So this is going to be equal to 10 minus 2X multiplied by 10 minus 2 X. OK. Multiplied by X, that's our volume, and notice here that our volume is a function of X. So if we expand our function here, then our volume as a function of X is going to be equal to 100 X minus 40 X2 + 4 X cubed. No, to maximize the volume, let's differentiate VF X with respect to X and set it to 0. So maximizing the effects. That is, finding the first derivative and setting it to 0. OK, then. Notice that The first derivative of VFX to find it we can differentiate, and here we are going to differentiate each term using the polar rule. The derivative of 100 X is 100, of -40 X squared is 80X, and of 4 X cubed is 12 x 2. So this first derivative, or this is the first derivative, and no setting it to 0. Then we're going to have 100 minus 80 X plus 12 x squared equal to 0. Now we can try to solve here for X, OK. We can divide through by 4, which means that it's gonna become 25 minus 20 X 2 plus 3 X2 equals 0. And now by quadratic factorization, we could rewrite this as 3X minus 5 multiplied by 3X minus 5, and that equals 0. Now if we think about it here. That means in this case, the, well, 3 x minus 5 multiplied by X minus 5, my apologies, will be equal to 0. And in that case, then X is either going to be 5/3 of a foot. R X is going to be 5 ft. But now, remember, if we go back to our dimensions, we know the length and the width are 10 minus 2X. If X equals 5, let's write that as a note here, if X equals 5 ft, then that would be implying that the length is going to be equal to 10 minus 2 multiplied by 5, and that's 0. But that's not possible because the length of our box cannot be 0 ft. That means our box would not exist. Therefore, X equals 5 ft is not a possible value. Only X equals 5/3 of a foot would be a possible value for the critical point. But no, we need to test if it corresponds to a maximum, OK. So let's test if X equals 5/3 of a foot. Corresponds to a maximum. And we can use, we can test it using the 2nd derivative test. Basically, what the second derivative test tells us is that if our second derivative is negative, the function is concave down, which means that the value corresponds to a maximum. But if it's positive, then the function is concave up and corresponds to a minimum. So here if we take the second derivative of VX, which means that we're going to differentiate the first derivative, OK. So that's gonna be the derivative of 100 minus 80 X plus 12 X squared. And that derivative is gonna be equal to -80 plus 24 X again applying the polar rule for these terms. And no At X equals 5/3, OK. Then the second derivative. It's gonna be equal to -80 plus 24 multiplied by 5/3. Which is gonna be -80 + 40, which equals -40. Since our second derivative is negative. OK. Since our second derivative is negative, that means the function is concave on. Which then tells us that X equals 5/3 of a foot corresponds. OK, let me write that down here, corresponds to a maximum. So now we have the value of X that that maximizes our value. OK? And now that we have the maximum or the value of X to maximize the volume, then we can find the maximum volume by substituting that value of X. So now, this means then. The maximum value of Z and X is 5/3 of a foot is gonna be equal to 100 multiplied by 5/3. We're putting it back in the original function. That's minus 40 multiplied by 5/3 squared, OK, plus 4 multiplied by 5/3 cubed. Which is gonna be equal to 2000 divided by 27 cubic feet. Therefore, the maximum, the dimension of X. That maximizes our volume is 5/3 of a foot and the maximum volume is 2000 divided by 27 cubic feet. Thanks a lot for watching everyone. I hope this video helped.

Folded boxes

a. Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 5 ft by 8 ft. The resulting piece of cardboard is then folded into a box without a lid. Find the volume of the largest box that can be formed in this way.

Key Concepts

Volume of a Box

Optimization

Derivatives

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