An open box is made from a rectangular piece of cardboard measuring 16 cm by 10cm. Four equal squares are to be cut from each corner and flaps folded up.
Find the length of the side of the square which makes the volume of the box as large as possible.
Find the largest volume
because x was cut off each corner. Draw a diagram, and you will see that when the corners are turned up, the sides have 2x cut off.
Well, let's think outside the box for a moment. Or in this case, inside the box. We have a rectangular piece of cardboard and we want to make an open box out of it by cutting squares from the corners. To maximize the volume, we need to find the perfect square size.
Now, if we imagine folding up the flaps, we'll end up with a box that has dimensions of length = 16 - 2x, width = 10 - 2x, and height = x. The volume of a rectangular box is calculated by multiplying the length, width, and height. So, our task is to maximize the volume function.
V(x) = (16 - 2x)(10 - 2x)(x)
Now, instead of being a square, let's calculate this function and find the x value that gives us the maximum volume. Don't worry, I won't go full mathematical clown on you... maybe just a little bit.
To simplify things, let's just expand the equation and bring it to life:
V(x) = 4x³ - 52x² + 160x
Now, to find the maximum volume, we need to find the derivative of this function and set it equal to zero. This will give us the critical points, or the potential maximums and minimums.
dV(x)/dx = 12x² - 104x + 160 = 0
Let's solve this equation. Hang on tight, it's going to get bumpy... just like a clown car ride!
Using the quadratic formula, you'd stumble across the values x = 5/3 and x = 10/3. Now, let's get real and throw away x = 5/3 because it is not within the range of our cardboard piece dimensions.
So, the winning value is x = 10/3. It might be a bit odd, but hey, clowns are all about being quirky!
To find the largest volume, just plug this winning value back into our original volume function:
V(10/3) = (16 - 2(10/3))(10 - 2(10/3))(10/3)
Simplifying it further (I won't torture you anymore, I promise), the largest volume is approximately 151.11 cm³.
So, the perfect square size for the largest volume is 10/3 cm, and the largest volume itself is 151.11 cm³. Let there be boxes!
To find the length of the side of the square that makes the volume of the box as large as possible, we can start by visualizing the process.
1. Start with a rectangular piece of cardboard measuring 16 cm by 10 cm.
2. Cut out a square from each corner of the cardboard. Let's call the side length of the square "x". After cutting the squares, the dimensions of the remaining cardboard will be (16-2x) cm by (10-2x) cm.
3. Fold up the flaps to form an open box. The height of the box will be "x" cm.
Now, to find the largest possible volume, we need to maximize the dimensions of the box. The volume of a rectangular box is given by the formula:
Volume = length * width * height
Substituting the dimensions from step 2 and the height from step 3, we have:
Volume = (16-2x) * (10-2x) * x
To find the largest volume, we can find the critical points of this equation by taking the derivative with respect to "x" and setting it equal to zero:
d/dx (Volume) = 0
Let's differentiate the equation:
d/dx [(16-2x) * (10-2x) * x] = 0
Expanding and simplifying:
(20-4x) * (10-2x) + (16-2x) * (10-2x) = 0
Now, solve this equation for "x" to find the length of the side of the square that maximizes the volume.
Calculating this equation may be a bit complex or time-consuming. We can use a graphing calculator or software to find the value of "x" that satisfies the equation.
Alternatively, we can substitute different values of "x" and calculate the corresponding volumes until we find the largest one. Starting with reasonable values (such as x = 1, 2, 3, etc.) and comparing the volumes, we can gradually approach the maximum volume.
the base of the box is 16-2x by 10-2x, and the height is x
v = x(16-2x)(10-2x)
v'= 4(x-2)(3x-20)
20/3 >5, so it's too large
so, max volume occurs when x=2
The box is 6x12x2 with volume 144
@steve
Why is the base of the box 16-2x by 10-2x
(where did the 2x come from?)