# Maximum Volume Hard

1. Nov 5, 2005

### dagg3r

Maximum Volume Hard!!!

hey guys got this maximum volume question so bloody hard !!! i got most of the steps right but got stuck on the last section (V)

1.
A closed box has a surface area of 18 square metres. let the dimensions of the three sides x,y,z and let volume be V
(i) write the volume V in terms of x,y,z
aight i got this right:
V=xyz
(ii) use result that surface area is 18 s metrs, find z in terms of x and y
1 got
Z = (9-xy)/(x+y)
(iii)
Find V in terms of x and y with results from (i) and (ii)

V = (9xy - x^2y^2) / (x+y)

(iv) using reults in (iii) find dv/dx and dv/dy where "d" represents the "day symbol"
i got
dv/dx = (9y^2-x^2y-2xy^3) / ((x+y)^2)
dv/dy = (9x^2-x^2y-2x^3y) / ((x+y)^2)

(v) for values of x,y,z to get max volume, i tried to let dv/dx and dv/dy = 0 no luck at all please help!!! thanks

2. Nov 5, 2005

### HallsofIvy

You have a slight error in the derivatives which may be a typo:
$$\frac{\partial V}{\partial x}= \frac{9y^2- x^2y^2- 2xy^3}{(x+y)^2}$$
$$\frac{\partial V}{\partial y}= \frac{9x^2- x^2y^2- 2x^2y}{(x+y)^2}$$
Notice the middle term is x2y2. The symmetry should have been a clue that it was not x2y in both!
Set each equation equal to 0, Multiply both sides by (x+y)2, divide the first equation by y2 and the second by x2 to get 9- x2- 2xy=0 and 9-y2- 2xy= 0. Subtract one equation from the other to eliminate the "9" and "2xy" terms. Remember that x and y, being measurements, must both be positive. The answer shouldn't surprise you.

3. Nov 5, 2005

### benorin

Note that since $$V(x,y)=V(y,x)$$, it is of necessity that $$x=y$$ at all extrema.

4. Nov 5, 2005

### HallsofIvy

That does not follow automatically! Certainly, if certain value of x gives an extremum (for some y) we could use symmetry to argue that that same value of y must give an extremum (for some x). But for a general function of x and y it is possible that, say, (1, 3) and (3, 1) give extrema.

(You are of course, completely correct for this problem- in fact, this could be done trivially by noting that the volume is symmetric in x, y, and z.

5. Nov 5, 2005

### benorin

Not only is the function to be max/minimized symmetric in its variables, so is the constraint. I must admit, I hadn't considered that a function which was symmetric in its variables (along with given contraints in the same variables) could give extrema that were not symmetric in those variables. Could you given an example? Or just enough to make me slap my forehead?

-Ben