Answer :
To minimize the cost of manufacturing a rectangular box with a square base and open top, we need to determine the dimensions such that the total cost is minimized. Let’s proceed step-by-step:
### Step 1: Define variables
- Let [tex]\( x \)[/tex] be the length of one side of the square base (in cm).
- Let [tex]\( h \)[/tex] be the height of the box (in cm).
### Step 2: Volume constraint
The volume [tex]\( V \)[/tex] of the box is given by:
[tex]\[ V = x^2 \cdot h = 670 \][/tex]
Thus, we can express [tex]\( h \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ h = \frac{670}{x^2} \][/tex]
### Step 3: Cost calculations
The total cost is the sum of the cost of the base and the cost of the sides.
- Base cost: The area of the base is [tex]\( x^2 \)[/tex] square centimeters.
- Cost of the base: [tex]\( \text{cost\_base} = 0.90 \cdot x^2 \)[/tex]
- Sides cost: There are 4 sides, each with an area of [tex]\( x \cdot h \)[/tex] square centimeters.
- Total side area: [tex]\( 4 \cdot (x \cdot h) = 4 \cdot x \cdot \frac{670}{x^2} = \frac{2680}{x} \)[/tex]
- Cost of the sides: [tex]\( \text{cost\_sides} = 0.90 \cdot \frac{2680}{x} \)[/tex]
Thus, the total cost function [tex]\( C(x) \)[/tex] is:
[tex]\[ C(x) = 0.90 \cdot x^2 + 0.90 \cdot \frac{2680}{x} \][/tex]
### Step 4: Optimize the cost function
To find the value of [tex]\( x \)[/tex] that minimizes the cost, we need to find the critical points by differentiating [tex]\( C(x) \)[/tex] with respect to [tex]\( x \)[/tex] and setting the derivative to zero.
Differentiate [tex]\( C(x) \)[/tex]:
[tex]\[ C'(x) = \frac{d}{dx} \left( 0.90 \cdot x^2 + 0.90 \cdot \frac{2680}{x} \right) \][/tex]
[tex]\[ C'(x) = 0.90 \cdot 2x - 0.90 \cdot \frac{2680}{x^2} \][/tex]
[tex]\[ C'(x) = 1.80x - 0.90 \cdot \frac{2680}{x^2} \][/tex]
[tex]\[ C'(x) = 1.80x - \frac{2412}{x^2} \][/tex]
Set [tex]\( C'(x) = 0 \)[/tex] to find the critical points:
[tex]\[ 1.80x - \frac{2412}{x^2} = 0 \][/tex]
[tex]\[ 1.80x^3 = 2412 \][/tex]
[tex]\[ x^3 = \frac{2412}{1.80} = 1340 \][/tex]
[tex]\[ x = \sqrt[3]{1340} \][/tex]
Calculate [tex]\( x \)[/tex]:
[tex]\[ x \approx \sqrt[3]{1340} \approx 11.02 \, \text{cm} \][/tex]
### Step 5: Calculate [tex]\( h \)[/tex] and verify dimensions
Using [tex]\( x = 11.02 \)[/tex] cm, find [tex]\( h \)[/tex]:
[tex]\[ h = \frac{670}{(11.02)^2} \approx \frac{670}{121.44} \approx 5.52 \, \text{cm} \][/tex]
### Step 6: Calculate the minimum cost
Substitute [tex]\( x \)[/tex] and [tex]\( h \)[/tex] back into the cost function:
[tex]\[ \text{Cost} = 0.90 \cdot (11.02)^2 + 0.90 \cdot \frac{2680}{11.02} \][/tex]
[tex]\[ \text{Cost} \approx 0.90 \cdot 121.44 + 0.90 \cdot 243.25 \][/tex]
[tex]\[ \text{Cost} \approx 109.30 + 218.93 \][/tex]
[tex]\[ \text{Cost} \approx 328.23 \][/tex]
### Final Answer
The dimensions of the box that minimize the cost are:
- Side of the base [tex]\( x \approx 11.02 \)[/tex] cm
- Height [tex]\( h \approx 5.52 \)[/tex] cm
The minimum cost of manufacturing the box, rounded to the nearest cent, is:
[tex]\[ \boxed{$328.23} \][/tex]
### Step 1: Define variables
- Let [tex]\( x \)[/tex] be the length of one side of the square base (in cm).
- Let [tex]\( h \)[/tex] be the height of the box (in cm).
### Step 2: Volume constraint
The volume [tex]\( V \)[/tex] of the box is given by:
[tex]\[ V = x^2 \cdot h = 670 \][/tex]
Thus, we can express [tex]\( h \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ h = \frac{670}{x^2} \][/tex]
### Step 3: Cost calculations
The total cost is the sum of the cost of the base and the cost of the sides.
- Base cost: The area of the base is [tex]\( x^2 \)[/tex] square centimeters.
- Cost of the base: [tex]\( \text{cost\_base} = 0.90 \cdot x^2 \)[/tex]
- Sides cost: There are 4 sides, each with an area of [tex]\( x \cdot h \)[/tex] square centimeters.
- Total side area: [tex]\( 4 \cdot (x \cdot h) = 4 \cdot x \cdot \frac{670}{x^2} = \frac{2680}{x} \)[/tex]
- Cost of the sides: [tex]\( \text{cost\_sides} = 0.90 \cdot \frac{2680}{x} \)[/tex]
Thus, the total cost function [tex]\( C(x) \)[/tex] is:
[tex]\[ C(x) = 0.90 \cdot x^2 + 0.90 \cdot \frac{2680}{x} \][/tex]
### Step 4: Optimize the cost function
To find the value of [tex]\( x \)[/tex] that minimizes the cost, we need to find the critical points by differentiating [tex]\( C(x) \)[/tex] with respect to [tex]\( x \)[/tex] and setting the derivative to zero.
Differentiate [tex]\( C(x) \)[/tex]:
[tex]\[ C'(x) = \frac{d}{dx} \left( 0.90 \cdot x^2 + 0.90 \cdot \frac{2680}{x} \right) \][/tex]
[tex]\[ C'(x) = 0.90 \cdot 2x - 0.90 \cdot \frac{2680}{x^2} \][/tex]
[tex]\[ C'(x) = 1.80x - 0.90 \cdot \frac{2680}{x^2} \][/tex]
[tex]\[ C'(x) = 1.80x - \frac{2412}{x^2} \][/tex]
Set [tex]\( C'(x) = 0 \)[/tex] to find the critical points:
[tex]\[ 1.80x - \frac{2412}{x^2} = 0 \][/tex]
[tex]\[ 1.80x^3 = 2412 \][/tex]
[tex]\[ x^3 = \frac{2412}{1.80} = 1340 \][/tex]
[tex]\[ x = \sqrt[3]{1340} \][/tex]
Calculate [tex]\( x \)[/tex]:
[tex]\[ x \approx \sqrt[3]{1340} \approx 11.02 \, \text{cm} \][/tex]
### Step 5: Calculate [tex]\( h \)[/tex] and verify dimensions
Using [tex]\( x = 11.02 \)[/tex] cm, find [tex]\( h \)[/tex]:
[tex]\[ h = \frac{670}{(11.02)^2} \approx \frac{670}{121.44} \approx 5.52 \, \text{cm} \][/tex]
### Step 6: Calculate the minimum cost
Substitute [tex]\( x \)[/tex] and [tex]\( h \)[/tex] back into the cost function:
[tex]\[ \text{Cost} = 0.90 \cdot (11.02)^2 + 0.90 \cdot \frac{2680}{11.02} \][/tex]
[tex]\[ \text{Cost} \approx 0.90 \cdot 121.44 + 0.90 \cdot 243.25 \][/tex]
[tex]\[ \text{Cost} \approx 109.30 + 218.93 \][/tex]
[tex]\[ \text{Cost} \approx 328.23 \][/tex]
### Final Answer
The dimensions of the box that minimize the cost are:
- Side of the base [tex]\( x \approx 11.02 \)[/tex] cm
- Height [tex]\( h \approx 5.52 \)[/tex] cm
The minimum cost of manufacturing the box, rounded to the nearest cent, is:
[tex]\[ \boxed{$328.23} \][/tex]
