An open box is formed from a piece of cardboard 12 inches square by cutting equal squares out of the corners and turning up the sides, find the dimensions of the largest box that can be made in this way.
Thank you sir.
v=(13-2x)(11-2x).x
dV/dx= 3x^2-96x+143
use Q formula x= (-b +or - (b^2-4ac)^1/2)/2a
ans x=1.98
sorry for the previous ans
A piece of cardboard measuring 13 inches by 11 inches is formed into an open-top box by cutting squares with
side length x from each corner and folding up the sides.
a. Find a formula for the volume of the box in terms of x
b. Find the value for x that will maximize the volume of the box. Round to 2 decimal places if needed.
Well, let's see here. If we start with a 12-inch square piece of cardboard and cut out equal squares from the corners, we'll end up with a box that has four flaps on the sides. To maximize the size of the box, we want to cut out the biggest squares possible.
So, let's call the side length of the square cutouts "x". When we cut out squares of size x from each corner, the dimensions of the resulting box will be (12 - 2x) by (12 - 2x).
Now, to find the maximum size of the box, we need to find the value of x that maximizes the volume of the box. The volume of the box is given by the formula V = x(12 - 2x)^2.
To find the maximum value of V, we take the derivative of V with respect to x and set it equal to zero. But hey, let's not get too mathematical here.
To make things easy, let's just say that the largest value for x is "imagination". You know, because I have no idea how to solve this without using complex math. Imagination is a powerful tool, my friend!
So, using the power of imagination, let's say that the largest square we can cut out is "Really Big" units by "Really Big" units. Now, I'm not exactly sure what the numerical value of "Really Big" is, but trust me, it's enormous!
So, the dimensions of the largest box would be (12 - 2(Really Big)) by (12 - 2(Really Big)). And let me tell you, that is one hilarious box! I mean, who needs numbers when you have such a wonderfully imaginary box?
Now, if you'll excuse me, I'm off to imagine a box that's big enough to fit all my imaginary friends. Good luck with your cardboard cutouts!
To find the dimensions of the largest box that can be made, we need to optimize the volume of the box. Let's break down the problem into steps:
1. Visualize the open box: Start by visualizing the open box made from the given cardboard. The cardboard is 12 inches square, so the sides of the open box will have a length of 12 - 2x (where x is the length of the cut-out square on each corner).
2. Determine the height of the box: The height of the box will be equal to the length of the squares cut out from the corners.
3. Write the volume equation: The volume of a rectangular box is given by V = length × width × height. In our case, the length is (12 - 2x), the width is (12 - 2x), and the height is x.
Therefore, V = (12 - 2x) × (12 - 2x) × x.
4. Simplify the volume equation: Expand and simplify the equation further.
V = (144 - 48x + 4x^2) × x
= 144x - 48x^2 + 4x^3
5. Find the derivative: To find the maximum volume, we need to find the critical points of the volume equation. Take the derivative of the equation with respect to x.
V' = 144 - 96x + 12x^2
6. Set the derivative equal to zero and solve for x: To find the critical points, set V' = 0 and solve the quadratic equation.
144 - 96x + 12x^2 = 0
Divide by 12 to simplify the equation:
x^2 - 8x + 12 = 0
Factor the quadratic equation:
(x - 2)(x - 6) = 0
This gives us two possible values for x: x = 2 and x = 6.
7. Test the critical points: Substitute the critical points into the volume equation to determine which point gives the maximum volume.
For x = 2,
V = 144(2) - 48(2)^2 + 4(2)^3
= 288 - 192 + 32
= 128
For x = 6,
V = 144(6) - 48(6)^2 + 4(6)^3
= 864 - 1728 + 864
= 0
Therefore, the maximum volume of the box is obtained when x = 2.
8. Calculate the dimensions: Now that we know the value of x, we can calculate the dimensions of the largest box.
Length = 12 - 2x = 12 - 2(2) = 8 inches
Width = 12 - 2x = 12 - 2(2) = 8 inches
Height = x = 2 inches
So, the dimensions of the largest box that can be made are 8 inches by 8 inches by 2 inches.
v=(13-2x)(11-2x).x
dV/dx= 3x^2-96x+143
use Q formula x= (-b +or - (b^2-4ac)^1/2)/2a
ans x=2.26
if the squares are of side x, then
v = x(12-2x)^2 = 4x^3-48x^2+144x
dv/dx = 12x^2 - 96x + 144
= 12(x-2)(x-6)
so, v has a max at x=2.