A parcel delivery service a package only of the length plus girth (distance around) does not exceed 24 inches.
A) Find the dimensions of a rectangular box with square ends that satisfies the delivery service's restriction and has a 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 a maximum volume. What is the maximum volume?
Let the size of the square (cross-section) be s.
Then we need to maximize
V=s²(24-2s) with respect to s.
First find the derivative and equate to zero:
dV/ds = 48s-6s²=0
means
s=0 or s=8
s=0 corresponds to a minimum volume and
s=8 corresponds to a maximum volume.
So the maximum volume is given by
8x8x8 inches, as is evident by intuition.
but an 8x8x8 box has length+girth = 8+32 = 40 inches, so it will not work.
We need to optimize s^2(24-4s) since a square has 4 sides.
v = 24s^2 - 4s^3
v' = 48s - 12s^2
v'=0 when s=4
So, a 4x4x8 box has max volume.
Do (B) similarly
Good catch, Steve.
Thank you.
To solve both questions, we can use the concept of optimization. We want to find the dimensions that maximize the volume of the package while satisfying the delivery service's restriction.
A) Rectangular Box with Square Ends:
1. Let's assume the length of the rectangular box is L, and each side of the square ends is x.
2. The girth is the sum of the lengths of the four sides, excluding the square ends. In this case, the girth is 2L.
3. Based on the given restriction, 2L + 4x (since there are 4 sides of length x) should not exceed 24 inches, so we have 2L + 4x ≤ 24.
4. To maximize the volume, we need to find the maximum value of V = L * x * x (length times width times height).
To find the maximum volume, we can use the method of substitution:
1. Rearrange the inequality from step 3 to solve for L: L ≤ (24 - 4x) / 2.
2. Substitute this value of L into the volume equation: V = ((24 - 4x) / 2) * x * x.
3. Simplify the equation to one variable: V = (12 - 2x) * (x^2).
4. Expand the equation: V = 12x^2 - 2x^3.
5. To maximize the volume, we need to find the critical points, which are the points where the derivative of V with respect to x is equal to zero.
dV/dx = 24x - 6x^2 = 0.
6. Solve for x: 24x = 6x^2.
Divide both sides by 6x (assuming x is nonzero): 4 = x.
7. Since x represents the side length of the square ends, x = 4.
8. Substitute this value of x back into the volume equation to find the maximum volume: V = (12 - 2(4)) * (4^2) = 8 * 16 = 128 cubic inches.
Therefore, the maximum volume of the rectangular box with square ends is 128 cubic inches, and its dimensions are: length = (24 - 4x) / 2 = (24 - 4(4)) / 2 = 8 inches, square sides = x = 4 inches.
B) Cylindrical Container:
1. Let's assume the radius of the cylinder is r, and the height is h.
2. The girth of the cylindrical container is the circumference of the circular ends, which is 2πr.
3. The length is the height of the cylinder, h.
4. Based on the given restriction, 2πr + h should not exceed 24 inches, so we have 2πr + h ≤ 24.
5. To maximize the volume, we need to find the maximum value of V = πr^2h (pi times radius squared times height).
To find the maximum volume, we can use the method of substitution:
1. Rearrange the inequality from step 4 to solve for h: h ≤ 24 - 2πr.
2. Then substitute this value of h into the volume equation: V = πr^2(24 - 2πr).
3. Simplify the equation to one variable: V = 24πr^2 - 2πr^3.
4. To maximize the volume, we need to find the critical points: dV/dr = 48πr - 6πr^2 = 0.
5. Solve for r: 48πr = 6πr^2.
Divide both sides by 6πr (assuming r is nonzero): 8 = r.
6. Since r represents the radius of the cylinder, r = 8.
7. Substitute this value of r back into the volume equation to find the maximum volume: V = π(8^2)(24 - 2π(8)) = 64π(24 - 16π) cubic inches.
8. Use approximations to get a numerical value for the maximum volume.
Therefore, the maximum volume of the cylindrical container is approximately 1505.7 cubic inches, and its dimensions are: radius = r = 8 inches, height = 24 - 2πr = 24 - 2π(8) = 24 - 16π inches.