Chapter 4.6, Problem 14E

To determine

To calculate: Thecost of material for the cheapest such container if rectangular storage container with an open top to have volume of $10{\text{ m}}^3$ and length of its base is twice the width. Material for the base costs $\$10$ per square meter. Material for sides costs $\$6$ per square meter.

Answer to Problem 14E

The cost of material for the cheapest such container is $\text{\$163}\text{.54(approx)}$

Explanation of Solution

Given information:

A rectangular storage container with an open top is to have volume of $10{\text{ m}}^3$ with length of its base is twice the width. The base costs of material is $\$10$ per square meter and the sides costs of material is $\$6$ per square meter.

Formula used:

Let $l$ be the length $w$ be the width and $h$ be the height of cuboid.

I. The lateral surface area of cuboid $A=2wh+2hl$

II. Volume of open box $V=lwh$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

a) $x=a$ is a point of local maximum value of $f$, if

i) $f^{\prime }(a)=0$ and

ii) $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima

b) $x=a$ is a point of local maximum value of $f$, if

i) $f^{\prime }(a)=0$ and

ii) $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

c) $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

d)

i. If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$

ii. If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

The rectangular storage container with an open top is to have volume of $10{\text{ m}}^3$ with length of its base is twice the width

  $\begin{array}{l}\text{Let width}=w\text{Then,}\\ \text{ lenght}\left(l\right)=2w\end{array}$

Recall that, Let $l$ be the length $w$ be the width and $h$ be the height of cuboid.

Volume of open box $V=lwh$

Substitute the values, and solve for $h$ :

  $\begin{array}{l}2w\times w\times h=10\\ w^2\times h=5\\ h=\frac{5}{w^2}\mathrm{......}\left(1\right)\end{array}$

Let $l$ be the length $w$ be the width and $h$ be the height of cuboid

.

The lateral surface area of cuboid $A=2wh+2hl$

The base costs of material is $\$10$ per square meter and the sides costs of material is $\$6$ per square meter.

  $\begin{array}{l}\text{Total cost}\left(C\right)=10\times \text{Area of base of cuboid +6}\times \text{Area of lateral surface area of cuboid}\\ \text{ =10}\times lw+6\times \left(2wh+2hl\right)\end{array}$

Substitute $l=2w\text{ and }h=\frac{5}{w^2}$ and further simplify,

  $\begin{array}{l}{C}^{\prime }\left(w\right)=120.14\ge 0\text{at Total cost}\left(C\right)\text{=10}\times lw+6\times \left(2wh+2hl\right)\\ =10\times 2w\times w+6\times 2\times w\times \frac{5}{w^2}+6\times 2\times \frac{5}{w^2}\times 2w\\ =20w^2+\frac{60}{w}+\frac{120}{w}\\ \text{Total cost}\left(C\right)\text{=10}\times lw+6\times \left(2wh+2hl\right)\\ =10\times 2w\times w+6\times 2\times w\times \frac{5}{w^2}+6\times 2\times \frac{5}{w^2}\times 2w\\ =20w^2+\frac{60}{w}+\frac{120}{w}\\ C\left(w\right)=20w^2+\frac{180}{w}\end{array}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

a) $x=a$ is a point of local maximum value of $f$, if

i. $f^{\prime }(a)=0$ and

ii. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima

b) $x=a$ is a point of local maximum value of $f$, if

i. $f^{\prime }(a)=0$ and

ii. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

c) $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

d)

If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$

If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $\begin{array}{l}{C}^{\prime }\left(w\right)=20\times 2w-\frac{180}{w^2}\\ C^{\prime }\left(w\right)=40w-\frac{180}{w^2}\mathrm{......}\left(2\right)\end{array}$

Solve for $C^{\prime }\left(w\right)=0$ ,

  $\begin{array}{l}40w-\frac{180}{w^2}=0\\ 40w=\frac{180}{w^2}\\ w^3=\frac{180}{40}\\ w=\sqrt[3]{\frac{9}{2}}\\ w=1.65(\text{approx})\end{array}$

Differentiate equation $\left(2\right)$ with respect to $w$

  $\begin{array}{l}{C}^{\prime }\left(w\right)=40+2\times \frac{180}{w^3}\\ C^{\prime }\left(w\right)=40+\frac{360}{w^3}\end{array}$

Substitute $w=1.65(\text{approx})$ and simplified

  $\begin{array}{l}{C}^{\prime }\left(1.65\right)=40+2\times \frac{180}{{\left(1.65\right)}^3}\\ C^{\prime }\left(w\right)=40+\frac{360}{1.65\times 1.65\times 1.65}\\ C^{\prime }\left(w\right)=40+80.14\\ C^{\prime }\left(w\right)=120.14(\text{appox})\end{array}$

Therefore,

  $C^{\prime }\left(w\right)=120.14\ge 0\text{at }w=1.65$

For $w=1.65$ volume of container is minimize

Substitute $w=1.65$ in equation $(1)$

  $\begin{array}{l}h=\frac{5}{{\left(1.65\right)}^2}\\ h=\frac{5}{1.65\times 1.65}\\ h=1.84(\text{approx})\end{array}$

And

  $\begin{array}{l}l=\frac{10}{wh}\\ l=\frac{10}{1.65\times 1.84}\\ l=3.29(\text{approx})\end{array}$

  $\begin{array}{l}\text{The total cost}\left(C\right)\text{=10}\times lw+6\times \left(2wh+2hl\right)\\ \text{ =10}\times 3.29\times 1.65+6\times \left(2\times 1.65\times 1.84+2\times 1.84\times 3.29\right)\\ \text{ =163}\text{.54(approx)}\end{array}$

The cost of material for the cheapest such container is $\text{\$163}\text{.54(approx)}$