Respuesta :
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.
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 [tex]\boxed{13{\text{ in}}}{\text{ and }}\boxed{{\text{6}}{\text{.5 in}}}[/tex] and the minimum surface area is [tex]\boxed{507{\text{ i}}{{\text{n}}^2}}.[/tex]
Further explanation:
Given:
The volume of the rectangular box is [tex]62.5{\text{ i}}{{\text{n}}^3}[/tex]
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,
[tex]\boxed{{\text{Surface Area}} = 4xy + {x^2}}[/tex]
The volume of the box is [tex]62.5{\text{ i}}{{\text{n}}^3}.[/tex]
[tex]\begin{aligned}{\text{Volume}}&= {x^2}y\\\frac{{62.5}}{{{x^2}}}&= y\\\end{aligned}[/tex]
The surface area of the box can be expressed as,
[tex]\begin{aligned}{\text{Surface area}}&= {x^2} + 4xy\\&= {x^2} + 4x \times \frac{{62.5}}{{{x^2}}}\\&= {x^2} + \frac{{250}}{x}\\\end{aligned}[/tex]
Differentiate the surface area with respect to “x”.
[tex]\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}[/tex]
Substitute 0 for [tex]\dfrac{d}{{dx}}\left( {{\text{Surface area}}}\right)[/tex] in above equation to obtain the value of x.
[tex]\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}[/tex]
Further solve the above equation.
[tex]\begin{aligned}{x^3}&= \frac{{250}}{2}\\{x^3}&= 125\\x&= \sqrt[3]{{125}}\\x&= 5\\\end{aligned}[/tex]
The side of the base is [tex]5{\text{ in}}.[/tex]
The height of the box can be obtained as follows,
[tex]\begin{aligned}y&= \frac{{62.5}}{{{x^2}}}\\&= \frac{{62.5}}{{{5^2}}}\\&=\frac{{62.5}}{{25}}\\&= 2.5\\\end{alligned}[/tex]
The height of the box is [tex]y = 2.5{\text{ in}}.[/tex]
The surface area of the box can be calculated as follows,
[tex]\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}[/tex]
The dimensions of minimize surface area are [tex]\boxed{5{\text{ in}}}{\text{ and }}\boxed{{\text{2}}{\text{.5 in}}}[/tex] and the minimum surface area is [tex]\boxed{75{\text{ i}}{{\text{n}}^2}}.[/tex]
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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.
