Final answer:
To maximize the volume of a box made from a 10in by 12in cardboard sheet by removing squares from each corner and folding the sides, we use calculus optimization techniques. A value for x that maximizes volume is found by deriving the volume function, setting the derivative to zero, solving for x, and then determining the maximum with the second derivative test.
Step-by-step explanation:
To find the dimensions of the box with the maximum volume that can be made from a 10in by 12in sheet of cardboard by removing a square from each corner and folding up the sides, we can use optimization in calculus.
Let the side of the square cut from each corner be x inches. After cutting out the squares and folding up the sides, the dimensions of the box will be: length (12 - 2x), width (10 - 2x), and height x, making the volume V = (12 - 2x)(10 - 2x)x.
Now, we need to find the value of x that maximizes volume. To do this, we take the derivative of the volume with respect to x, set it to zero, and solve for x. This will give us the critical points, and we can use the second derivative test to find the maximum. Once x is found, we plug it back into the dimensions to find the size of the box.
It's important to note that x must be in the range of 0 to 5 inches since the length and width cannot be negative, and in practical terms x can't be zero because you'd have no box at all.