An open-topped rectangular box is to be constructed from a 24 inch by 36 inch piece of cardboard by cutting
out squares of equal sides from the corners and then folding up the sides. What size squares should be cut out
of each of the corners in order to maximize the volume of the box and what is the maximum volume?
I tried this out by multiplying the width and the length together and got 896.
I then went ahead and plugged it into the formula for a rectangular prism with x as my variable => V(x)=x(896-2x)^2
Unfortunately, I realized that this would lead me to get a positive second derivative as an answer. I wouldn't get a max with that, rather, a min.
Not sure where I went wrong with this one.
new length = 36 - 2x
new width = 24 - 2 x
so
V = (36-2x)(24-2x)x
= 864x - 120 x^2 + 4 x^3
I am glad you do derivatives because I do not feel like finding the vertex.
dV/dx = 864 -240 x + 12 x^2
0 = 72 - 20 x + x^2
x = [ 20 +/- sqrt(400 -288) ]/2
= [ 20 +/- 10.6 ] /2
= 15.3 or 4.7
4.7 works
0 when x = 120/8 = 15
To find the size of the squares that should be cut out from each corner to maximize the volume of the box, you are on the right track. However, it seems there was an error in your calculation, and you need to differentiate the volume formula correctly to find the maximum.
Let's go through the steps again:
1. Start with a 24-inch by 36-inch rectangular piece of cardboard.
2. Cut out squares of equal sides, x inches, from each corner.
3. Fold up the sides to form the open-topped box.
To calculate the volume, we need to determine the dimensions of the base and height.
The base length of the box will be the original length minus two side lengths of the square cutouts, so it is given by: 36 - 2x.
The base width of the box will be the original width minus two side lengths of the square cutouts, so it is given by: 24 - 2x.
The height of the box will be the side length of the square cutouts, so it is given by: x.
Now, we can calculate the volume of the box:
Volume of the box, V = Length × Width × Height
= (36 - 2x) × (24 - 2x) × x
= x(36 - 2x)(24 - 2x)
To find the value of x that maximizes the volume, we need to differentiate V(x) with respect to x and set it equal to zero.
So let's differentiate:
V'(x) = (36 - 2x)(24 - 2x) + x(-4)(24 - 2x) + x(36 - 2x)(-4)
Simplifying, we get:
V'(x) = 4x³ - 120x² + 864x
To find the critical points, set V'(x) equal to zero and solve for x:
4x³ - 120x² + 864x = 0
Now you can solve this cubic equation to find the critical points. From there, you can determine the size of the squares to be cut out from the corners that will maximize the volume of the box.
If you have the values for x after solving the equation, please provide them so we can continue from there.
To solve this problem, we need to apply optimization techniques to find the maximum volume. Let's break it down step by step:
1. Let's assume that squares of side length 'x' are cut out from each corner of the rectangular cardboard.
2. By doing so, the length of the resulting box will be (36 - 2x) inches, and the width will be (24 - 2x) inches, while the height or depth will be 'x' inches.
3. Now, let's express the volume of the box in terms of x. The volume of the box is given by multiplying its length, width, and height: V(x) = (36 - 2x)(24 - 2x)x.
4. To find the maximum value of the volume, we need to take the derivative of V(x) with respect to x and set it equal to zero. This will give us the critical points.
5. Differentiating V(x) with respect to x: dV(x)/dx = (36 - 2x)(-2)(24 - 2x) + x(-4)(24 - 2x) = -4(3x - 36)(x - 12) - 8x(12 - x).
6. Simplifying the derivative: dV(x)/dx = -12(x - 12)(x - 3) - 8x(12 - x).
7. Setting the derivative equal to zero and solving for x: -12(x - 12)(x - 3) - 8x(12 - x) = 0.
8. Expanding and solving the equation: -12x^2 + 180x - 432 + 8x^2 - 96x = 0. Combining like terms: -4x^2 + 84x - 432 = 0.
9. Solving the quadratic equation either through factoring or using the quadratic formula, we find that x = 6 or x = 18.
10. We have two critical points, x = 6 and x = 18. Since the length and width of the cardboard are 36 and 24 inches, respectively, we can see that x cannot be greater than 12 (half of the smaller dimension).
11. Plugging the critical points into the volume equation, we find that V(6) = 6(24)(36 - 12) = 6(24)(24) = 34,560, and V(18) = 18(24)(0) = 0.
12. Therefore, the maximum volume is obtained when squares of side length 6 inches are cut out from each corner. The maximum volume is 34,560 cubic inches.
So, the correct answer is that squares of side length 6 inches should be cut out from each corner, and the maximum volume of the box is 34,560 cubic inches.