An open top box is to be constructed so that its base is twice as long as it is wide. Its volume is to be 2400cm cubed. How do you find the dimensions that will minimize the amount of cardboard requiblack?

valtricotinevh
2022-07-17
Answered

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Jamarion Roth

Answered 2022-07-18
Author has **13** answers

Secondary Equation : $w\times \left(2w\right)\times h=2400$

Solve for h:

$h=\frac{1200}{{w}^{2}}$

Primary Equation for Surface Area (with open top) :

$f\left(w\right)=2{w}^{2}+2wh+2\left(2w\right)h$

Now use the Secondary Equation by substituting for h:

$f\left(w\right)=2{w}^{2}+\left(2w\right)\left(\frac{1200}{{w}^{2}}\right)+\left(4w\right)\left(\frac{1200}{{w}^{2}}\right)$

Simplify:

$f\left(w\right)=2{w}^{2}+\frac{7200}{w}$

Now find the derivative and set equal to zero:

$f\prime =4w-\frac{7200}{{w}^{2}}=0$

${w}^{3}=1800$

$w={\left(1800\right)}^{\frac{1}{3}}\approx 12.1644$

[I'll leave it for you to prove that this is a minimum]

Finally, the dimensions that minimize the surface area are:

$w={\left(1800\right)}^{\frac{1}{3}}$

$l=2{\left(1800\right)}^{\frac{1}{3}}$

$h=\frac{1200}{{w}^{2}}=\frac{1200}{{\left(1800\right)}^{\frac{2}{3}}}$

Solve for h:

$h=\frac{1200}{{w}^{2}}$

Primary Equation for Surface Area (with open top) :

$f\left(w\right)=2{w}^{2}+2wh+2\left(2w\right)h$

Now use the Secondary Equation by substituting for h:

$f\left(w\right)=2{w}^{2}+\left(2w\right)\left(\frac{1200}{{w}^{2}}\right)+\left(4w\right)\left(\frac{1200}{{w}^{2}}\right)$

Simplify:

$f\left(w\right)=2{w}^{2}+\frac{7200}{w}$

Now find the derivative and set equal to zero:

$f\prime =4w-\frac{7200}{{w}^{2}}=0$

${w}^{3}=1800$

$w={\left(1800\right)}^{\frac{1}{3}}\approx 12.1644$

[I'll leave it for you to prove that this is a minimum]

Finally, the dimensions that minimize the surface area are:

$w={\left(1800\right)}^{\frac{1}{3}}$

$l=2{\left(1800\right)}^{\frac{1}{3}}$

$h=\frac{1200}{{w}^{2}}=\frac{1200}{{\left(1800\right)}^{\frac{2}{3}}}$

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Suppose my original constrained minimisation problem is

$\underset{x}{min}f(g(x),x)\phantom{\rule{1em}{0ex}}\text{s.t.}\phantom{\rule{1em}{0ex}}g(x)=3$

I would like to know if this equivalent to solving the unconstrained minimisation problem

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If not, when are these two problems equivalent?

Suppose my original constrained minimisation problem is

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I would like to know if this equivalent to solving the unconstrained minimisation problem

$\underset{x}{min}f(3,x)$

If not, when are these two problems equivalent?