Questions: A box with a rectangular base is to be constructed of material costing 2 / in^2 for the sides and bottom and 3 / in^2 for the top. If the box is to have volume 1,215 in^3 and the length of its base is to be twice its width, what dimensions of the box will minimize its cost of construction? What is the minimal cost?
Transcript text: A box with a rectangular base is to be constructed of material costing $\$ 2 / \mathrm{in}^{2}$ for the sides and bottom and $\$ 3 / \mathrm{in}^{2}$ for the top. If the box is to have volume $1,215 \mathrm{in}^{3}$ and the length of its base is to be twice its width, what dimensions of the box will minimize its cost of construction? What is the minimal cost?
Solution
Solution Steps
To solve this problem, we need to find the dimensions of the box that minimize the cost of construction while maintaining a fixed volume. The box has a rectangular base with the length being twice the width. We will express the cost as a function of the width and height, then use calculus to find the minimum cost.
Define variables for the width, length, and height of the box.
Use the given volume to express the height in terms of the width.
Write the cost function in terms of the width and height, considering the different costs for the sides, bottom, and top.
Use calculus to find the critical points of the cost function and determine the minimum cost.
To solve this problem, we need to find the dimensions of a box that minimize the cost of construction given the constraints. Let's break it down step by step.
Step 1: Define Variables and Constraints
Let:
\( w \) be the width of the base of the box (in inches),
\( l = 2w \) be the length of the base of the box (since the length is twice the width),
\( h \) be the height of the box (in inches).
The volume of the box is given by:
\[
V = l \cdot w \cdot h = 2w \cdot w \cdot h = 2w^2h = 1215
\]
From this, we can express \( h \) in terms of \( w \):
\[
h = \frac{1215}{2w^2}
\]
Step 2: Express the Cost Function
The cost of the box is determined by the surface area and the cost per square inch of the materials used. The surface area \( S \) of the box is:
Bottom: \( l \times w = 2w \times w = 2w^2 \)
Top: \( l \times w = 2w \times w = 2w^2 \)
Sides: \( 2(l \times h + w \times h) = 2(2wh + wh) = 6wh \)
The cost function \( C \) is:
\[
C = 2 \times 2w^2 + 3 \times 2w^2 + 2 \times 6wh
\]
\[
C = 4w^2 + 6w^2 + 12wh
\]
\[
C = 10w^2 + 12wh
\]
Substitute \( h = \frac{1215}{2w^2} \) into the cost function:
\[
C = 10w^2 + 12w \left(\frac{1215}{2w^2}\right)
\]
\[
C = 10w^2 + \frac{14580}{w}
\]
Step 3: Minimize the Cost Function
To find the minimum cost, we need to take the derivative of \( C \) with respect to \( w \) and set it to zero:
\[
\frac{dC}{dw} = 20w - \frac{14580}{w^2}
\]
Set the derivative equal to zero:
\[
20w - \frac{14580}{w^2} = 0
\]
\[
20w^3 = 14580
\]
\[
w^3 = \frac{14580}{20}
\]
\[
w^3 = 729
\]
\[
w = \sqrt[3]{729} = 9
\]
Step 4: Calculate Other Dimensions
Now that we have \( w = 9 \), we can find \( l \) and \( h \):
\[
l = 2w = 2 \times 9 = 18
\]
\[
h = \frac{1215}{2 \times 9^2} = \frac{1215}{162} = 7.5
\]
Step 5: Calculate the Minimal Cost
Substitute \( w = 9 \), \( l = 18 \), and \( h = 7.5 \) back into the cost function:
\[
C = 10 \times 9^2 + \frac{14580}{9}
\]
\[
C = 10 \times 81 + 1620
\]
\[
C = 810 + 1620 = 2430
\]
Final Answer
The dimensions of the box that minimize the cost are:
Width: \( w = 9 \) inches
Length: \( l = 18 \) inches
Height: \( h = 7.5 \) inches
The minimal cost of construction is:
\[
\boxed{2430}
\]