by cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, the cardboard may be turned into an open box. if the cardboard is 16 inches long and 10 inches wide, find the dimensions of the box that will yield the max volume. what is the max volume?
Two Congruent squares are removed from one end of a rectangle 10 inch by 20 inch piece of cardboard. Two congruent rectangles are removed from the other end Determine the value of x so that the resulting box has maximum volume
To find the dimensions of the box that will yield the maximum volume, we need to follow a step-by-step process. Let's break it down:
Step 1: Visualize the problem
Imagine a rectangular piece of cardboard with a length of 16 inches and a width of 10 inches. We will cut identical squares from each corner of the cardboard.
Step 2: Determine the dimensions of the cut squares
Let's say the side length of the square we cut from each corner is 'x' inches.
Step 3: Calculate the dimensions of the resulting open box
Since we cut the squares from each corner, the length and width of the box will be reduced by 2x (x inches from each of the four edges). Therefore, the new length of the box will be (16 - 2x) inches, and the new width will be (10 - 2x) inches. The height of the box will be 'x' inches (equal to the length of the cut squares).
Step 4: Define the volume function in terms of 'x'
Now, we can define the volume of the open box in terms of 'x'. The volume, V, is given by the product of length, width, and height: V = (16 - 2x)(10 - 2x)x.
Step 5: Find the value of 'x' that maximizes the volume
To find the value of 'x' that maximizes the volume, we take the derivative of the volume function with respect to 'x' and set it equal to zero. Then, solve the resulting equation to find critical points. Finally, determine if these critical points yield a maximum or minimum.
Step 6: Calculate the maximum volume
Once you find the value of 'x' that maximizes the volume, substitute it back into the volume function to calculate the actual maximum volume.
By going through these steps, you can solve the problem and find the dimensions of the box that will yield the maximum volume, as well as the maximum volume itself.
let each side of the square to be cut out be x inches
length of base = 16-2x
width of base = 10-2x
height of box = x inches
V = x(16-2x)(10-2x) , where 0 < x < 5
I would now expand that to get a cubic
take the derivative, which is a quadratic,
set the derivative equal to zero and solve for x
Very straight forward question, most Calculus texts use that question as an introduction to optimization.