The dimensions of minimize surface area are \(\boxed{13{\text{ in}}}{\text{ and }}\boxed{{\text{6}}{\text{.5 in}}}\) and the minimum surface area is \(\boxed{507{\text{ i}}{{\text{n}}^2}}.\)
Further explanation:
Given:
The volume of the rectangular box is \(62.5{\text{ i}}{{\text{n}}^3}\)
Explanation:
Consider the base length of the square box as “x”.
Consider the height of the box as “y”.
The surface area of the open box can be expressed as follows,
\(\boxed{{\text{Surface Area}} = 4xy + {x^2}}\)
The volume of the box is \(62.5{\text{ i}}{{\text{n}}^3}.\)
\(\begin{aligned}{\text{Volume}}&= {x^2}y\\\frac{{62.5}}{{{x^2}}} &= y\\\end{aligned}\)
The surface area of the box can be expressed as,
\(\begin{aligned}{\text{Surface area}} &= {x^2} + 4xy\\&= {x^2} + 4x \times \frac{{62.5}}{{{x^2}}}\\&= {x^2} + \frac{{250}}{x} \\\end{aligned}\)
Differentiate the surface area with respect to “x”.
\(\begin{aligned}\frac{d}{{dx}}\left( {{\text{Surface area}}} \right)&= \frac{d}{{dx}}\left( {{x^2} + \frac{{4394}}{x}} \right)\\&= 2x - \frac{{250}}{{{x^2}}}\\\end{aligned}\)
Substitute 0 for \(\dfrac{d}{{dx}}\left( {{\text{Surface area}}} \right)\) in above equation to obtain the value of x.
\(\begin{aligned}\frac{d}{{dx}}\left( {{\text{Surface area}}} \right)&= 0\\2x - \frac{{250}}{{{x^2}}}&= 0\\2x&=\frac{{250}}{{{x^2}}}\\x \times {x^2}&= \frac{{125}}{2}\\\end{aligned}\)
Further solve the above equation.
\(\begin{aligned}{x^3}&= \frac{{250}}{2}\\{x^3} &= 125\\x &= \sqrt[3]{{125}}\\x &= 5\\\end{aligned}\)
The side of the base is \(5{\text{ in}}.\)
The height of the box can be obtained as follows,
\(\begin{aligned}y&= \frac{{62.5}}{{{x^2}}}\\&= \frac{{62.5}}{{{5^2}}}\\&=\frac{{62.5}}{{25}}\\&= 2.5\\\end{aligned}\)
The height of the box is \(y = 2.5{\text{ in}}.\)
The surface area of the box can be calculated as follows,
\(\begin{aligned}{\text{Surface area}} &= {\left( 5 \right)^2} + 4\left( {5 \times 2.5} \right)\\&= 25 + 4\left( {12.5} \right)\\&= 25 + 50\\&= 75{\text{ i}}{{\text{n}}^2}\\\end{aligned}\)
The dimensions of minimize surface area are \(\boxed{5{\text{ in}}}{\text{ and }}\boxed{{\text{2}}{\text{.5 in}}}\) and the minimum surface area is \(\boxed{75{\text{ i}}{{\text{n}}^2}}.\)
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Application of Derivatives
Keywords: Drum, tight containers, open top, square base, volume of 62.5 inches cubed, rectangular box, minimum surface area, dimensions, designing.
