I would use your addition equation to solve for one parameter. Plug that equation into your product. Then what do you suppose you should do?
Another way to do this is to use "Lagrange multipliers".
You want to maximize f(a,b,c)= abc subject to the constraint that g(a,b,c)= 13a+ 5b+ 2c= 172, a constant. At a minimum or maximum point, the gradients will be parallel:
for some constant .
so that we have , , and where " " is the "Lagrange multiplier".
Since the precise value of is not necessary for the solution of Lagrange multiplier problems, I find it usually best to start by eliminating by dividing one equation by another. For example, dividing the first equation by the second, so that . Similarly, dividing the first equation by the third, so that .
Put those into the constraint, 13a+ 5b+ 2c= 172, to get an equation you can solve for a.
Listen to Ackbeet.
EDIT=You can also do it the way Hallsofivy showed above=EDIT
Write out the equations, solve for a variable and substitute. If your in calc three then thats the easy part. This is what you have:
you want to maximize , and you know that you want to express it in terms of the other equation (this ensures that you've "mixed" the two restricting equations in your problem). We also have:
Pick a or b and explicitly define one (in terms of b and c, or a and c; depending on which you pick) by moving around variables and numbers in the equaiton above. Substitute it into the P=ab equation, then simplify. I'm assuming since your in calc 3 your intructor might have shown you an example? Do you understand where to go next? Give the algebra a try first, post back if your truly still stuck
Just alittle add on here, incase you didnt know. But this is the higher dimensional analog to finding the critical points of a single variable function. This is how you find the critical points of a 3-dimensional surface. At critical points, there exist four possible occurances (from my understanding).
[1] There can be a minimum at a critical point
[2] There may be a maximum at a critical point
[3] There may be a saddle at a critical point (i.e. if you draw two perpendicular axes at this point, one axis would slope downward in both directions, and the other would slope upaward in both directions)
[4] There may be nothing that you can determine about the behavior of the graph at that critical point (i.e. if the determinant of the square matrix of partial derivitives is equal to zero)