A box with an open top is to be made from a square piece of cardboard by cutting equal squares from the corners and turning up the sides. If the piece of cardboard measures 12 in on the side, find the size of the squares that must be cut out to yield the maximum volume of the box.
So far i got V=x(12-2x)^2, then i took the derivative to be (12-2x)^2-4x(12-2x), then i end up with 12(x-6)(x-2)=0 which means x=6 and x=2.... but that doesn't make sense because each side is 12-2x and if x is 6 the side is 0. Any help you can provide would be appreciated!
yes, the zero derivative can be either a maximum or a minimum
If x = 6, we have the minimum, in fact zero
If x = 2, it seems we have a maximum which is the answer you want.
Oh okay, that makes sense. thank you!
To solve this problem, let's go step by step.
The first step was correct. You correctly identified that the volume, V, of the box as a function of x (the size of the squares cut out of the corners) is given by V = x(12-2x)^2.
Next, we need to find the value of x that yields the maximum volume. To do this, we need to take the derivative of V with respect to x and set it equal to zero. Let's find the derivative:
V' = dV/dx = (12-2x)^2 - 2x(12-2x) * d(12-2x)/dx
Simplifying this expression, we have:
V' = (12-2x)^2 - 2x(12-2x) * (-2)
V' = (12-2x)^2 + 4x(12-2x)
Expanding further:
V' = 144 - 48x + 4x^2 + 48x - 8x^2
V' = 144 - 8x^2 + 4x^2
V' = 144 - 4x^2
Now, set V' equal to zero and solve for x:
144 - 4x^2 = 0
Dividing both sides by 4:
36 - x^2 = 0
Rearranging:
x^2 = 36
Taking the square root of both sides:
x = 6 or x = -6
At this point, we have two possible solutions: x = 6 and x = -6. However, we are dealing with the size of squares, which must be a positive value. Therefore, we discard the negative solution, x = -6.
So, the only valid solution is x = 6.
This means that squares with dimensions 6 inches by 6 inches should be cut out from each corner of the 12-inch by 12-inch cardboard to maximize the volume of the box.