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If 2200 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Calculus1
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Use Lagrange multipliers to minimize \(f(s)=s^3\) subject to \(g(s)=5s^2=2200\). That is, set up the equation\[\Lambda(s,\lambda)=s^3+\lambda(5s^2-2200),\]equate its gradient to zero, and solve the resultant system for \(s\) and \(\lambda\):\[\begin{align}\nabla\Lambda&=0\\\Rightarrow\Lambda_s&=3s^2+\lambda(10s-2200)=0,\\\Lambda_\lambda&=5s^2-2200=0.\end{align}\]Can you do this?
I made a typo: \(\Lambda_s\) should be \(3s^2+10\lambda s\).
unfortunatly no, my professor hasn't introduced the Lagrange multipliers, nor lambda, usually uses x and y as variables

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Other answers:

really @across, legrange multipliers?
|dw:1351867483030:dw| The Volume of the box will be: \[V = l*w*h\] This is the equation you have to maximize, but first it must be simplified to take the derivative with one variable. One constraint is that the base is square, this means: \[l = w\] The other constraint is given by the total material you have to make the box: \[SA_{total}= 2200cm^2\]dfsfdffa Form an equation for SA and set it equal to this value, and solve for one variable \[SA_{total}=SA_{sides} + SA_{bottom}=2200\] \[4*l*h+l^2=2200\] \[h=\frac{2200-l^2}{4*l}\]
you can now make your substitutions into the original equation to get Volume wrt to one variable: \[V = l*w*h\] \[V = l*(l)*(\frac{2200-l^2}{4*l})\] \[V=\frac{l}{4}(2200-l^2)\] It's now a Max/Min question that you solve by by taking derivative of Volume wrt length and setting to zero: \[\frac{dV}{dl}=0\] but it all comes from the initial picture

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