Verified Answer :
Let the square base of the container be of side s inches and the height of the container be h inches, then
Surface are of the container, A = s^2 + 4sh
For minimum surface area, dA / ds + dA / dh = 0
i.e. 2s + 4h + 4s = 0
6s + 4h = 0
s = -2/3 h
But, volume of container = 62.5 in cubed
i.e. s^2 x h = 62.5
(-2/3 h)^2 x h = 62.5
4/9 h^2 x h = 62.5
4/9 h^3 = 62.5
h^3 = 62.5 x 9/4 = 140.625
h = cube root of (140.625) = 5.2 inches
s = 2/3 h = 3.47
Therefore, the dimensions of the square base of the container is 3.47 inches and the height is 5.2 inches.
The minimum surface area = s^2 + 4sh = (3.47)^2 + 4(3.47)(5.2) = 12.02 + 72.11 = 84.13 square inches.
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,
\(\begin{aligned}y&= \frac{{62.5}}{{{x^2}}}\\&= \frac{{62.5}}{{{5^2}}}\\&=\frac{{62.5}}{{25}}\\&= 2.5\\\end{alligned}\)
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 \(\boxed{75{\text{ i}}{{\text{n}}^2}}.\)
Learn more:
1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.
2. Learn more about equation of circle brainly.com/question/1506955.
3. Learn more about range and domain of the function https://brainly.com/question/3412497
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.