Find the dimensions of a box with a square base that maximizes volume given a su

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Question: Find the dimensions of a box with a square base that maximizes volume given a surface area of 600 sq. units. (2020)

Options:

Correct Answer: 12, 12

Exam Year: 2020

Solution:

Let x be the side of the base and h the height. The surface area constraint gives 2x^2 + 4xh = 600. Max volume occurs at x = 12.

Find the dimensions of a box with a square base that maximizes volume given a surface area of 600 sq. units. (2020)

Find the dimensions of a box with a square base that maximizes volume given a surface area of 600 sq. units. (2020)

Step 1: Understand that we need to find the dimensions of a box with a square base that has a maximum volume.
Step 2: Let 'x' be the length of one side of the square base and 'h' be the height of the box.
Step 3: Write down the formula for the surface area of the box. The surface area (SA) is given by SA = 2x^2 + 4xh.
Step 4: Set the surface area equal to 600 square units: 2x^2 + 4xh = 600.
Step 5: Rearrange the equation to express 'h' in terms of 'x': h = (600 - 2x^2) / (4x).
Step 6: Write the formula for the volume (V) of the box: V = x^2 * h.
Step 7: Substitute the expression for 'h' into the volume formula: V = x^2 * ((600 - 2x^2) / (4x)).
Step 8: Simplify the volume formula: V = (600x - 2x^3) / 4.
Step 9: To find the maximum volume, take the derivative of the volume function with respect to 'x' and set it to zero.
Step 10: Solve the equation to find the value of 'x' that maximizes the volume. This gives x = 12.
Step 11: Substitute x = 12 back into the equation for 'h' to find the height of the box.

Optimization – The problem involves maximizing the volume of a box under a given surface area constraint.
Surface Area and Volume Relationships – Understanding how to relate the dimensions of the box to its surface area and volume.
Calculus – Utilizing derivatives to find maximum values in optimization problems.