Question

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.

Answer 1

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?

Answer 2

I made a typo: \(\Lambda_s\) should be \(3s^2+10\lambda s\).

Answer 3

unfortunatly no, my professor hasn't introduced the Lagrange multipliers, nor lambda, usually uses x and y as variables

Answer 4

really @across, legrange multipliers?

Answer 5

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
\[SA_{total}=SA_{sides} + SA_{bottom}=2200\]
\[4*l*h+l^2=2200\]
\[h=\frac{2200-l^2}{4*l}\]