Question

Unit 7 - Optimization Problems

5. Maximize Volume - We have a piece of cardboard that is 14 inches by 10 inches and
Boy we're going to cut out the corners as shown below and fold up the sides to form a
0212 box, also shown below. Determine the height of the box that will give a maximum
que volume.

Answer 1

Explanation:

To maximize the volume of the box, we need to find the height that will maximize the volume of the box.

Let's start by finding an expression for the volume of the box. The box has dimensions of 14-2x by 10-2x by x, where x is the height of the box. The volume of the box is:

V(x) = (14-2x)(10-2x)(x)

Expanding this expression, we get:

V(x) = 4x^3 - 48x^2 + 140x

To find the value of x that maximizes this expression, we can take the derivative of V(x) with respect to x and set it equal to zero:

V'(x) = 12x^2 - 96x + 140 = 0

We can solve this quadratic equation using the quadratic formula:

x = [96 ± sqrt(96^2 - 4(12)(140))]/(2(12)) = [96 ± 16sqrt(6)]/24

We can simplify this to:

x = 4 ± sqrt(6)/3

Since the dimensions of the box must be positive, we can discard the negative solution:

x = 4 + sqrt(6)/3

So the height of the box that will give a maximum volume is approximately 5.61 inches (rounded to two decimal places).