A piece of cardboard measuring 11 inches by 12 inches is formed into an open-top box by cutting squares with side length
x from each corner and folding up the sides.
Find a formula for the volume of the box in terms of x
V(x)= ?
Find the value for x that will maximize the volume of the box
x = ?
see similar problems below, and a couple of posts just below your in the list.
To find the volume of the box in terms of x, we need to first determine the dimensions of the base of the box. When we cut squares with side length x from each corner and fold up the sides, the length of the base will be reduced by 2x (x on each side), and the width will be reduced by 2x as well. Therefore, the length of the base will be 11 - 2x and the width will be 12 - 2x.
The formula for the volume of a rectangular box is V = length * width * height. In this case, the height of the box will be x. So the formula for the volume of the box in terms of x is:
V(x) = (11 - 2x) * (12 - 2x) * x
To find the value of x that maximizes the volume of the box, we can take the derivative of V(x) with respect to x, set it equal to zero, and solve for x. Since this is a cubic equation, we will need to factor and solve.
Let's differentiate V(x):
V'(x) = [(11 - 2x) * (12 - 2x)]' * x + (11 - 2x) * (12 - 2x) * x'
Simplifying:
V'(x) = [(11 - 2x) * (12 - 2x)]' * x + (11 - 2x) * (12 - 2x)
V'(x) = [(132 - 46x + 4x^2)]' * x + 132 - 46x + 4x^2
V'(x) = (132 - 46x + 4x^2) * x' + 132 - 46x + 4x^2
V'(x) = (132 - 46x + 4x^2) + 132 - 46x + 4x^2
Simplifying further:
V'(x) = 264 - 92x + 8x^2
Now, set V'(x) equal to zero and solve:
264 - 92x + 8x^2 = 0
We can factor this equation:
8x^2 - 92x + 264 = 0
2x^2 - 23x + 66 = 0
Now we can solve for x by factoring or using the quadratic formula. In this case, the equation factors to:
(2x - 11)(x - 6) = 0
Setting each factor equal to zero:
2x - 11 = 0
x - 6 = 0
Solving for x:
2x = 11
x = 11/2
x = 5.5
and
x = 6
Thus, there are two critical points x = 5.5 and x = 6.
We can evaluate the volume of the box at each critical point to determine which one maximizes the volume. Plug x = 5.5 into V(x):
V(5.5) = (11 - 2(5.5)) * (12 - 2(5.5)) * 5.5
V(5.5) = 0 * 0 * 5.5
V(5.5) = 0
Now plug x = 6 into V(x):
V(6) = (11 - 2(6)) * (12 - 2(6)) * 6
V(6) = (11 - 12) * (12 - 12) * 6
V(6) = -1 * 0 * 6
V(6) = 0
Since both critical points yield a volume of 0, there is no maximum volume for this particular cardboard size and cutting method. The volume remains constant at 0 regardless of the value of x.