Questions: Part 1 of 5 HW Score: 16.67%, 5 of 30 points Points: 0 of 1 A parcel delivery service will deliver a package only if the length plus girth (distance around) does not exceed 60 inches. (A) Find the dimensions of a rectangular box with square ends that satisfies the delivery service's restriction and has maximum volume. What is the maximum volume? (B) Find the dimensions (radius and height) of a cylindrical container that meets the delivery service's restriction and has maximum volume. What is the maximum volume? (A) The dimensions of the rectangular box are in. (Use a comma to separate answers as needed.)

Solution Steps

To find the dimensions of the rectangular box with square ends that maximizes the volume under the delivery service's restriction, we start with the constraint:

\[ L + 4x = 60 \]

where \( L \) is the length and \( x \) is the side length of the square ends. Solving for \( L \):

\[ L = 60 - 4x \]

The volume \( V \) of the box is given by:

\[ V = x^2 L = x^2 (60 - 4x) = 60x^2 - 4x^3 \]

To find the critical points, we take the derivative of the volume with respect to \( x \):

\[ \frac{dV}{dx} = 120x - 12x^2 \]

Setting the derivative equal to zero:

\[ 120x - 12x^2 = 0 \implies 12x(10 - x) = 0 \]

This gives us critical points at \( x = 0 \) and \( x = 10 \). Since \( x = 0 \) does not provide a valid box dimension, we take \( x = 10 \).

Substituting \( x = 10 \) back into the equation for \( L \):

\[ L = 60 - 4(10) = 20 \]

Thus, the dimensions of the rectangular box are \( 10 \, \text{in} \) (side length) and \( 20 \, \text{in} \) (length).

The maximum volume of the rectangular box can be calculated as follows:

\[ V_{\text{max}} = 10^2 \cdot 20 = 2000 \, \text{in}^3 \]

Next, we find the dimensions of the cylindrical container that maximizes the volume under the same delivery service restriction. The constraint is:

\[ h + 2\pi r = 60 \]

Solving for \( h \):

\[ h = 60 - 2\pi r \]

The volume \( V \) of the cylinder is given by:

\[ V = \pi r^2 h = \pi r^2 (60 - 2\pi r) = 60\pi r^2 - 2\pi^2 r^3 \]

Taking the derivative with respect to \( r \):

\[ \frac{dV}{dr} = 120\pi r - 6\pi^2 r^2 \]

Setting the derivative equal to zero:

\[ 120\pi r - 6\pi^2 r^2 = 0 \implies 6\pi r(20 - \pi r) = 0 \]

This gives us critical points at \( r = 0 \) and \( r = \frac{20}{\pi} \). We take \( r \approx 6.3662 \).

Substituting \( r \) back into the equation for \( h \):

\[ h = 60 - 2\pi(6.3662) \approx 60 - 39.998 \approx 20.002 \]

The maximum volume of the cylindrical container can be calculated as follows:

\[ V_{\text{max}} \approx \pi (6.3662)^2 (20.002) \approx 2546.4812 \, \text{in}^3 \]

Final Answer