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?

Ariel Wilkinson
2022-09-06
Answered

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beshrewd6g

Answered 2022-09-07
Author has **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)

asked 2022-05-09

Most statements of Constraint Qualification I have found in the literature mention a locally "locally optimal solution" of the problem:

$\{\begin{array}{l}minf(x)\\ \text{s.t.}\\ {g}_{i}(x)\le 0\end{array}$

It is stated that when a C.Q. holds at a local optimum, then there exist Lagrange multipliers that satisfy KKT conditions.

But, I cannot get my head around this notion of local optimality. Does it mean locally optimal for the unconstrained problem? Does not local optimality imply the satisfaction of the KKT conditions?

$\{\begin{array}{l}minf(x)\\ \text{s.t.}\\ {g}_{i}(x)\le 0\end{array}$

It is stated that when a C.Q. holds at a local optimum, then there exist Lagrange multipliers that satisfy KKT conditions.

But, I cannot get my head around this notion of local optimality. Does it mean locally optimal for the unconstrained problem? Does not local optimality imply the satisfaction of the KKT conditions?

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Find global extrema of $f(x,y)=\frac{{x}^{2}}{2}+3{y}^{2}$ on the set $M(g):=\{(x,y)\in {\mathbb{R}}^{2}\mid g(x,y)=0\}$ where $g(x,y)={x}^{2}+{y}^{4}-25$.

$M(g)$ is compact and $f$ continuous so we know that there must exists a global maximum and global minimum. The conditions of the Lagrange multiplier methods are satisfied and if we solve the equations that result from the Lagrange multiplier method we get the following points:

$(0,\sqrt{5}),(0,-\sqrt{5}),(5,0),(-5,0),(4,\sqrt{3}),(-4,\sqrt{3}),(4,-\sqrt{3}),(-4,-\sqrt{3}).$

and

$f(0,\sqrt{5})=15,f(0,-\sqrt{5})=15,\phantom{\rule{0ex}{0ex}}f(5,0)=\frac{25}{2},f(-5,0)=\frac{25}{2},\phantom{\rule{0ex}{0ex}}f(4,\sqrt{3})=17,f(-4,\sqrt{3})=17,f(4,-\sqrt{3})=17,f(-4,-\sqrt{3})=17.$

So far so good.

However our sample solution says that "from the above values we see that $\frac{25}{2}$ is the global minimum and 17 the global maximum. "

As far as I have understood the Lagrange multiplier method it only delivers a necessary but not sufficient condition. So we don't know if one of the three points $\frac{25}{2},15,17$ is a saddle point. To make sure that the points are indeed extrema we have to resort to another method (e.g. plug in the condition into $f$).

Am I am right or is there something I don't see or didn't understand correctly?

$M(g)$ is compact and $f$ continuous so we know that there must exists a global maximum and global minimum. The conditions of the Lagrange multiplier methods are satisfied and if we solve the equations that result from the Lagrange multiplier method we get the following points:

$(0,\sqrt{5}),(0,-\sqrt{5}),(5,0),(-5,0),(4,\sqrt{3}),(-4,\sqrt{3}),(4,-\sqrt{3}),(-4,-\sqrt{3}).$

and

$f(0,\sqrt{5})=15,f(0,-\sqrt{5})=15,\phantom{\rule{0ex}{0ex}}f(5,0)=\frac{25}{2},f(-5,0)=\frac{25}{2},\phantom{\rule{0ex}{0ex}}f(4,\sqrt{3})=17,f(-4,\sqrt{3})=17,f(4,-\sqrt{3})=17,f(-4,-\sqrt{3})=17.$

So far so good.

However our sample solution says that "from the above values we see that $\frac{25}{2}$ is the global minimum and 17 the global maximum. "

As far as I have understood the Lagrange multiplier method it only delivers a necessary but not sufficient condition. So we don't know if one of the three points $\frac{25}{2},15,17$ is a saddle point. To make sure that the points are indeed extrema we have to resort to another method (e.g. plug in the condition into $f$).

Am I am right or is there something I don't see or didn't understand correctly?