What is the maximum volume in cubic inches of an open box to be made from a 12-inch by 16-inch piece of cardboard by cutting out squares of equal sides from the four corners and bending up the sides? your work must include a statement of the function and its derivative. give one decimal place in your final answer.

step-by-step explanation:

the volume of a box is: v = l * w * h. we are starting with a piece of cardboard 10 inches by 20 inches. if we let x represent the height of the box, x will also be the side of the square taken out of each corner.

the length of the box becomes 20 - 2x. the width of the box becomes 10 - 2x. the height of the box is x.

substituting in the equation for the volume gives:

v = x * (20 - 2x) * (10 - 2x)

multiply the right side to get:

v(x) = 4x3 - 60x2 + 200x

take the first derivative of v:

v'(x) = 12x2 - 120x + 200

to find x, set the first derivative equal to 0:

12x2 - 120x + 200 = 0

use the quadratic formula to solve for x:

a = 12

b = -120

c = 200

x = (120 ± √)2 - 4 * 12 * 200) / (2 * 12)

use the positive value first:

x = (120 + √(14400 - 9600) / 24

x = (120 + 69.3) / 24

x = 189.3 / 24

x = 7.9 inches

this solution can be discarded because it would make the width a negative number.

use the negative value next:

x = (120 - √(14400 - 9600) / 24

x = (120 - 69.3) / 24

x = 50.7 / 24

x = 2.1 inches

take the second derivative of v to check for a local maximum.

v''(x) = 24x - 120

substitute for x:

v''(2.1) = 24 * 2.1 - 120

the second derivative is less than zero, x is a maximum.

sides of the box:

length = 20 - 2 * 2.1 = 17.8 inches

width = 10 - 2 * 2.1 = 7.8 inches

height = 2.1 inches

step-by-step explanation:

in mathematics, a constraint is a condition of an optimization problem that the solution must satisfy. there are several types of constraints—primarily equality constraints, inequality constraints, and integer constraints. the set of candidate solutions that satisfy all constraints is called the feasible set.

hoped i'm right lol