1. ## Question about maximizing area.

A closed rectangle container with a square base must have a volume of 2250m^3. The material for the top and the sides of the container will cost $3/m^2. The material for teh bottom of the container will cost$2/m^2. Determine the dimensions of the box which will minimize the cost of the material for the box.

Now I know to find mins you take the 1st deriv test, but how do I apply this to this question?

2. Surface area is: $3(\underbrace{xy}_{\text{top}}+\underbrace{2xz+2yz }_{\text{sides}})+\underbrace{2xy}_{\text{bottom}}$

Therefore, we want $S=5xy+6xz+6yz$........[1]

subject to the constraint

$xyz=2250$.....[2]

Solve [1] for z and sub into [2]:

$z=\frac{2250}{xy}$

$S=5xy+6x(\frac{2250}{xy})+6y(\frac{2250}{xy})$

$=5xy+\frac{13500}{y}+\frac{13500}{x}$....[3]

Differentiate [3] with respect to x and y:

$S_{x}=5y-\frac{13500}{x^{2}}=0$...[4]

$S_{y}=5x-\frac{13500}{y^{2}}=0$...[5]

Solve [4] for y and sub into [5]:

$y=\frac{2700}{x^{2}}$

$5x-\frac{13500}{(\frac{2700}{x^{2}})^{2}}=5x-\frac{x^{4}}{540}=0$

Solving this for x, we find $x=3\cdot{10^{\frac{2}{3}}}\approx{13.92}$

Then $y=3\cdot{10^{\frac{2}{3}}}\approx{13.92}$

and $z=\frac{5\cdot{10^{\frac{2}{3}}}}{2}\approx{11.60}$

Check to see if it is, indeed a minimum. Use the Second Partials Test.

$D=S_{xx}(x,y)S_{yy}(x,y)-S_{xy}^{2}(x,y)=75>0$

Also, $S_{xx}(x,y)=10>0$

Therefore, we have a relative minimum.

Now, LaGrange multipliers. See if we get the same thing.

Let $f(x,y,z)=5xy+6xz+6yz$

$g(x,y,z)=xyz$

${\nabla}f=(5y+6z)i+(5x+6z)j+(6x+6y)k$

${\nabla}g=yxi+xzj+xyk$

$(5y+6z)i+(5x+6z)j+(6x+6y)k={\lambda}(yzi+xzj+xyk)$

$5y+6z=yz{\lambda}\rightarrow{\lambda}=\frac{5y+6z} {yz}$...[1]
$5x+6z=xz{\lambda}\rightarrow{\lambda}=\frac{5x+6z} {xz}$...[2]
$6x+6y=xy{\lambda}\rightarrow{\lambda}=\frac{6x+6y} {xy}$...[3]

From [1] and [2], we get x=y

From [2] and [3], we get $z=\frac{5y}{6}$

Sub these into the constraint, xyz, and we get:

$\frac{5y^{3}}{6}=2250$

Solving for y, we get $x=y=3\cdot{10^{\frac{2}{3}}}$

and $z=\frac{5\cdot{10^{\frac{2}{3}}}}{2}$

Same as before. Good. It checks.