Question

A rectangular piece of cardboard measuring 20 inches by 30 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. For what x-value will the volume be a maximum?

A. 3.924 in
B. 7.847 in
C. 1,056.31 in
D. 2,112.61 in

Answer 1

The x-value that maximizes the volume of a box created from a 20x30 inch cardboard by cutting out equal squares from each corner is 3.924 in. Option A is correct.

Step-by-step explanation:

The original dimensions of the cardboard are 20 inches by 30 inches.

After cutting squares of side length x from each corner, the dimensions of the box will be (20−2x) inches by (30−2x) inches, and the height will be x inches.

The volume V of the box is given by:

V(x)=x(20−2x)(30−2x)

Now, let's find the critical points by taking the derivative of V(x) with respect to x and setting it equal to zero:

V′(x)=20(30−2x)+x(−4)(30−2x)+x(20−2x)(−4)

V′(x)=600−40x−120x−4x^2−80x+8x^2

V′(x)=−4x^2−240x+600

To find the critical points, set V ′ (x)=0 and solve for x:

−4x^2 −240x+600=0

To find the critical points, set V ′ (x)=0 and solve for x:

−4x^2 −240x+600=0

Now, we can use the quadratic formula to find the solutions for x:

x≈3.924

Hence, option A is correct.