Question

Drum tight containers is designing an open-top, square-based, rectangular box that will have a volume of 62.5 in^3. what dimension will minimize surface area? what is the minimum surface area?

Answer 1

Square means L=W
V=62.5=LWH

L=W so
V=62.5=HL^2
SA=2(L^2+2LH)

we have
V=62.5=HL^2
solve for H
divide both sides by L^2
62.5/L^2=H
sub that for H in other equation

SA=2(L^2+2L(62.5/L^2))
SA=2(L^2+125/L)
SA=2L^2+250/L
find minimum of 2L^2+250L^-1
take the derivitive
4L-250L^-2, or 2(2L^3-125)/L^2
find where it equals zero
it equals zero at L=2.5∛4
L=W
if we evaluate 2L^2+250/L at L=2.5∛4, the value is 75∛2

H=62.5/L^2
H=$\frac{25 \sqrt[3]{2} }{4}$



dimentions are
L=2.5∛4
W=2.5∛4
H=$\frac{25 \sqrt[3]{2} }{4}$
minimum surface area is 75∛2 in^2 or aprox 94.4941 in^2

Answer 2

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 $\boxed{75{\text{ i}}{{\text{n}}^2}}.$

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 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.