Ariel Wilkinson

2022-09-06

A box with a square base and open top must have a volume of 32,000cm^3. How do you find the dimensions of the box that minimize the amount of material used?

beshrewd6g

Beginner2022-09-07Added 12 answers

The Volume of a box with a square base x by x cm and height h cm is $V={x}^{2}h$

The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.

The surface area of the box described is $A={x}^{2}+4xh$

We need A as a function of x alone, so we'll use the fact that $V={x}^{2}h=32,000$ cm^3

which gives us $h=\frac{32,000}{{x}^{2}}$, so the area becomes:

$A={x}^{2}+4x\left(\frac{32,000}{{x}^{2}}\right)={x}^{2}+\frac{128,000}{x}$

We want to minimize A, so

$A\prime =2x-\frac{128,000}{{x}^{2}}=0$ when $\frac{2{x}^{3}-128,000}{{x}^{2}}=0$

Which occurs when ${x}^{3}-64,000=0$ or x=40

The only critical number is x=40 cm.

The second derivative test verifies that A has a minimum at this critical number:$A\prime \prime =2+\frac{256,000}{{x}^{3}}$ which is positive at x=40.

The box should have base 40 cm by 40 cm and height 20 cm.

(use $h=\frac{32,000}{{x}^{2}}$ and x=40)

The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.

The surface area of the box described is $A={x}^{2}+4xh$

We need A as a function of x alone, so we'll use the fact that $V={x}^{2}h=32,000$ cm^3

which gives us $h=\frac{32,000}{{x}^{2}}$, so the area becomes:

$A={x}^{2}+4x\left(\frac{32,000}{{x}^{2}}\right)={x}^{2}+\frac{128,000}{x}$

We want to minimize A, so

$A\prime =2x-\frac{128,000}{{x}^{2}}=0$ when $\frac{2{x}^{3}-128,000}{{x}^{2}}=0$

Which occurs when ${x}^{3}-64,000=0$ or x=40

The only critical number is x=40 cm.

The second derivative test verifies that A has a minimum at this critical number:$A\prime \prime =2+\frac{256,000}{{x}^{3}}$ which is positive at x=40.

The box should have base 40 cm by 40 cm and height 20 cm.

(use $h=\frac{32,000}{{x}^{2}}$ and x=40)

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2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constantLet $f:{\mathbb{R}}^{2}\to \mathbb{R}$ be defined as

$f(x,y)=\{\begin{array}{ll}({x}^{2}+{y}^{2})\mathrm{cos}\frac{1}{\sqrt{{x}^{2}+{y}^{2}}},& \text{for}(x,y)\ne (0,0)\\ 0,& \text{for}(x,y)=(0,0)\end{array}$

then check whether its differentiable and also whether its partial derivatives ie ${f}_{x},{f}_{y}$ are continuous at $(0,0)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for ${f}_{x}$ as function is symmetric in $y$ and $x$. So ${f}_{x}$ turns out to be

${f}_{x}(x,y)=2x\mathrm{cos}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\mathrm{sin}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)$

which is definitely not $0$ as $(x,y)\to (0,0)$. Same can be stated for ${f}_{y}$. But how to proceed with the first part?