a box with no top is to be built by taking a 12''-by-16'' sheet of cardboard and cutting x-inch squares our of each corner and folding up the sides. find the value of x that maximizes the volume of the box
If you cut out corner squares with length x , the height of the box is x and the base dimensions are (12-2x) * (16 - 2x)
Box volume V = x(12-2x)(16-2x)
(0 < x < 6)
V = 192x -56x^2 + 4x^3
Set the derivative dV/dx = 0 and solve for x
A box with no top is to be built by taking a 12 in -by 16-in sheet of cardboard, cutting x-inch square out of each corner and finding up the sides. Find the value of x that maximize the volume of the box
Well, isn't that an interesting puzzle! So, we have a 12'' by 16'' sheet of cardboard, and we're going to cut out some squares from the corners to build a box with no top. Our job is to find the value of x that maximizes the volume of the box. Let's put on our thinking caps!
First, let's visualize what we're dealing with. When we cut out the squares from the corners and fold up the sides, the length of the box will correspond to the length of the original cardboard, which is 16'' minus twice the value of x (since we're cutting out two corners). Similarly, the width of the box will be 12'' minus twice the value of x.
So, the formula for the volume of this box is V = (16-2x)(12-2x)(x). Now, we want to find the value of x that maximizes this volume. How can we do that?
One way is to find the derivative of the volume function with respect to x and set it to zero. However, that might get a bit mathematical, and we'd rather keep things light-hearted, right?
Let's think outside the box! Or in this case, inside the box. Imagine that we have a really, really big x value. If x becomes larger than half the length or half the width, we'll end up with negative dimensions for the box, which doesn't make any sense. We can't have a negative length or width!
So, we know that x can't be larger than half the minimum of the length and the width. In our case, x can't be larger than half of 12 inches. So, x ≤ 6 inches.
To keep things symmetrical, let's also say that x ≤ 8 inches, which is half of 16 inches.
Now, can we find the maximum volume by trying different x values between 0 and 8 inches? Of course not! That sounds boring!
Instead, we can use the powers of our imagination and see that the maximum volume will likely occur at the maximum value of x, which is 8 inches in our case. This is because, as x increases, the length and width of the box decrease, and the volume is maximized when they are as small as possible.
Therefore, the value of x that maximizes the volume of the box is 8 inches.
Now, grab your cardboard, your scissors, and your lovely sense of humor, and get folding!
To find the value of x that maximizes the volume of the box, we can follow these steps:
1. Visualize the problem: Imagine the 12''-by-16'' sheet of cardboard. Four squares with side length x will be cut out from each corner, leaving four flaps to fold up and create the open box.
2. Determine the dimensions: Let's assume that after cutting the squares, the height of the box is h, and the side lengths of the square base are (12-2x) and (16-2x). So, the volume of the box, V, is given by V = (12-2x)(16-2x)h.
3. Express the volume in terms of a single variable: Since the height is not provided, we need to express h in terms of x. In this case, let's assume that the height of the box will be x to maximize the volume.
4. Calculate the volume: Substitute h = x into the volume equation, V = (12-2x)(16-2x)x, and simplify it to get V = 4x(12-x)(16-x).
5. Determine the critical points: To find the value of x that maximizes the volume, we need to find the critical points of the volume equation. These occur when the derivative of V with respect to x equals zero or when there is an undefined point. Let's calculate the derivative of V with respect to x.
V' = 4(12-x)(16-x) + 4x[-1(16-x) - 1(12-x)]
= 4(144 - 28x + x^2) - 4x(28 - 2x - 12 + x)
= 4(144 - 28x + x^2) - 4x(-2x + 16)
= 4(144 - 28x + x^2) + 8x^2 - 64x
= 4x^2 - 116x + 576.
Setting V' equal to zero and solving for x gives:
4x^2 - 116x + 576 = 0.
We can solve this quadratic equation to find the critical points.
6. Find the value of x: By factoring or using the quadratic formula, we can find that the two solutions are x = 6 and x = 24/7 ≈ 3.43.
7. Determine the maximum volume: We have found the critical points, but we need to determine which one maximizes the volume. To do this, we can compare the values of V at both critical points and at endpoints (where V is defined). Calculate the volume at x = 6, x = 24/7, x = 0, and x = 6.
V(6) = 4(6)(12-6)(16-6) = 4(6)(6)(10) = 1440.
V(24/7) = 4(24/7)(12-(24/7))(16-(24/7)) ≈ 330.14.
V(0) = 4(0)(12-0)(16-0) = 0.
V(6) = 4(6)(12-6)(16-6) = 1440.
Comparing these values, we can see that the maximum volume occurs at x = 6.
Therefore, the value of x that maximizes the volume of the box is 6 inches.