By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 16 in. long and 10 in. wide, find the dimensions of the box that will yield the maximum volume. (Round your answers to two decimal places.)
in (smallest value)
in
in (largest value)
Now I did the one below this one. I think you can do this one if you read that one carefully
remember length = 16 - 2 h
and width = 10 - 2 h
To find the dimensions of the box that will yield the maximum volume, we need to optimize the volume function with respect to the dimensions of the box.
Let's denote the length of the cut squares as x. Since we are cutting squares from each corner, the length of the resulting box will be (16 - 2x) inches and the width will be (10 - 2x) inches.
The height of the box will be x inches, as this is the length of the cut squares that we folded up to form the box.
Now, we can write the formula for the volume of the box:
V(x) = (16 - 2x)(10 - 2x)(x)
To find the maximum volume, we need to find the value of x that maximizes the volume function V(x).
To do this, we can take the derivative of the volume function V(x) with respect to x, set it equal to zero, and solve for x. The critical point we find will correspond to the maximum volume.
Let's find the derivative of V(x):
V'(x) = -4(x - 5)(x - 8)
Setting V'(x) equal to zero:
-4(x - 5)(x - 8) = 0
This equation gives us two critical points for x: x = 5 and x = 8.
Now, to determine which critical point gives the maximum volume, we can evaluate the volume function at each critical point and the endpoints of the feasible interval.
Evaluate V(x) at x = 5:
V(5) = (16 - 2(5))(10 - 2(5))(5) = 60 cubic inches
Evaluate V(x) at x = 8:
V(8) = (16 - 2(8))(10 - 2(8))(8) = 96 cubic inches
As for the endpoints of the feasible interval, we know that the length of the cut squares, x, must be less than half of the length and width of the cardboard. So the feasible interval for x is (0, 5) since if x ≥ 5, the resulting dimensions of the box will become negative or zero.
Evaluate V(x) at x = 0:
V(0) = (16 - 2(0))(10 - 2(0))(0) = 0 cubic inches
Evaluate V(x) at x = 5:
V(5) = (16 - 2(5))(10 - 2(5))(5) = 60 cubic inches
Therefore, the maximum volume of the box is 96 cubic inches, which occurs when x = 8.
The dimensions of the box that yield the maximum volume are:
Length: 16 - 2x = 16 - 2(8) = 0 inches
Width: 10 - 2x = 10 - 2(8) = -6 inches (not valid)
Height: x = 8 inches
So, the box has a maximum volume of 96 cubic inches with dimensions of 0 inches by 8 inches by 8 inches.