Answer :
To determine the dimensions that yield the least amount of material for a box with a square base, an open top, and a given volume, we need to minimize the surface area given the volume constraint. Here's the step-by-step approach:
1. Define the Variables:
- Let [tex]\( x \)[/tex] be the side length of the square base.
- Let [tex]\( h \)[/tex] be the height of the box.
2. Volume Constraint:
- The volume [tex]\( V \)[/tex] of the box is given by:
[tex]\[ V = x^2 h \][/tex]
- Given the volume [tex]\( V \)[/tex] is 13.5 cubic feet:
[tex]\[ x^2 h = 13.5 \][/tex]
3. Surface Area:
- The surface area [tex]\( A \)[/tex] of the box includes:
- The base: [tex]\( x^2 \)[/tex]
- The four sides: [tex]\( 4xh \)[/tex]
- Therefore, the surface area [tex]\( A \)[/tex] is:
[tex]\[ A = x^2 + 4xh \][/tex]
4. Express [tex]\( h \)[/tex] in Terms of [tex]\( x \)[/tex]:
- From the volume constraint, solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{13.5}{x^2} \][/tex]
5. Substitute [tex]\( h \)[/tex] into the Surface Area Equation:
- Substitute [tex]\( h = \frac{13.5}{x^2} \)[/tex] into the surface area equation:
[tex]\[ A = x^2 + 4x \left( \frac{13.5}{x^2} \right) \][/tex]
- Simplify the equation:
[tex]\[ A = x^2 + \frac{54}{x} \][/tex]
6. Minimize the Surface Area:
- To find the minimum surface area, take the derivative of [tex]\( A \)[/tex] with respect to [tex]\( x \)[/tex] and set it to zero:
[tex]\[ \frac{dA}{dx} = 2x - \frac{54}{x^2} \][/tex]
[tex]\[ \frac{dA}{dx} = 2x - 54x^{-2} \][/tex]
- Set the derivative equal to zero to find critical points:
[tex]\[ 2x - 54x^{-2} = 0 \][/tex]
[tex]\[ 2x = \frac{54}{x^2} \][/tex]
[tex]\[ 2x^3 = 54 \][/tex]
[tex]\[ x^3 = 27 \][/tex]
[tex]\[ x = 3 \][/tex]
7. Determine the Corresponding [tex]\( h \)[/tex]:
- Use [tex]\( x = 3 \)[/tex] in the volume constraint equation to find [tex]\( h \)[/tex]:
[tex]\[ h = \frac{13.5}{x^2} = \frac{13.5}{3^2} = \frac{13.5}{9} = 1.5 \][/tex]
8. Conclusion:
- The dimensions that minimize the amount of material used are:
- Side length of the base [tex]\( x = 3 \)[/tex] feet
- Height [tex]\( h = 1.5 \)[/tex] feet
Thus, the box with dimensions 3 feet by 3 feet for the base and 1.5 feet for the height will yield the least amount of material.
1. Define the Variables:
- Let [tex]\( x \)[/tex] be the side length of the square base.
- Let [tex]\( h \)[/tex] be the height of the box.
2. Volume Constraint:
- The volume [tex]\( V \)[/tex] of the box is given by:
[tex]\[ V = x^2 h \][/tex]
- Given the volume [tex]\( V \)[/tex] is 13.5 cubic feet:
[tex]\[ x^2 h = 13.5 \][/tex]
3. Surface Area:
- The surface area [tex]\( A \)[/tex] of the box includes:
- The base: [tex]\( x^2 \)[/tex]
- The four sides: [tex]\( 4xh \)[/tex]
- Therefore, the surface area [tex]\( A \)[/tex] is:
[tex]\[ A = x^2 + 4xh \][/tex]
4. Express [tex]\( h \)[/tex] in Terms of [tex]\( x \)[/tex]:
- From the volume constraint, solve for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{13.5}{x^2} \][/tex]
5. Substitute [tex]\( h \)[/tex] into the Surface Area Equation:
- Substitute [tex]\( h = \frac{13.5}{x^2} \)[/tex] into the surface area equation:
[tex]\[ A = x^2 + 4x \left( \frac{13.5}{x^2} \right) \][/tex]
- Simplify the equation:
[tex]\[ A = x^2 + \frac{54}{x} \][/tex]
6. Minimize the Surface Area:
- To find the minimum surface area, take the derivative of [tex]\( A \)[/tex] with respect to [tex]\( x \)[/tex] and set it to zero:
[tex]\[ \frac{dA}{dx} = 2x - \frac{54}{x^2} \][/tex]
[tex]\[ \frac{dA}{dx} = 2x - 54x^{-2} \][/tex]
- Set the derivative equal to zero to find critical points:
[tex]\[ 2x - 54x^{-2} = 0 \][/tex]
[tex]\[ 2x = \frac{54}{x^2} \][/tex]
[tex]\[ 2x^3 = 54 \][/tex]
[tex]\[ x^3 = 27 \][/tex]
[tex]\[ x = 3 \][/tex]
7. Determine the Corresponding [tex]\( h \)[/tex]:
- Use [tex]\( x = 3 \)[/tex] in the volume constraint equation to find [tex]\( h \)[/tex]:
[tex]\[ h = \frac{13.5}{x^2} = \frac{13.5}{3^2} = \frac{13.5}{9} = 1.5 \][/tex]
8. Conclusion:
- The dimensions that minimize the amount of material used are:
- Side length of the base [tex]\( x = 3 \)[/tex] feet
- Height [tex]\( h = 1.5 \)[/tex] feet
Thus, the box with dimensions 3 feet by 3 feet for the base and 1.5 feet for the height will yield the least amount of material.
