Finite Math Word Problem I have been having trouble with this word problem for a while. 
A bicycle manufacturer builds one-, three-, and ten-speed models. The bicycles need both aluminum and steel. The company has available 50,075 units of steel and 62,160 units of aluminum. The one,-three, and ten speed models need, respectively, 10, 20, 25 units of steel and 24, 12, 30, units of aluminum. How many of each type of bicycle should be made in order to maximize profit if the company makes 2 dollars per one speed, 4 dollars per three speed, and 10 dollars per ten-speed. What is the maximum possible profit.
Company Aluminum = 62,160
Company Steel = 50,075
One-Speed Bicycle = 2 dollars, 10 Steel, 24 Aluminum.
Three-speed Bicycle = 4 dollars, 20 Steel, 12 Aluminum.
Ten-speed Bicycle = 10 dollars, 25 Steel, 30 Aluminum.
I would like to know how many of each type must be made to maximize profits, and what the maxiumum profit is.
 A: Just to complement Ross' answer, I present the LPP formulation:
Let $x,y,z$ be the total number of one, three and ten speed bicycles manufactured respectively. 
\begin{align}
\max_{x,y,z}\quad &2x+4y+10z\\
10x+20y+25z&\leq 50075\\
24x+12y+30z&\leq 62160\\
x,y,z &\geq 0
\end{align}

You can solve this using the Simplex Method. 
If you aren't interested in the solution methodology and just want the answer, many commercial packages have the ability to solve such problems.
For instance, you can solve this in MATLAB using the following commands:
c = [-2 -4 -10]
A = [10 20 25; 24 12 30]
b = [50075;62160]
linprog(c,A,b,[],[],[0;0;0])

which gives you the same answer as the one Ross described. 
A: There are "official" techniques for linear programming problems like this, but with small numbers of equations one can do a lot by hand.  We have $50,075$ units of steel and $62,160$ units of aluminum.  The products have the characteristics: $$\begin {array} {}&1&3&10 \\ \hline Steel&10&20&25\\Aluminum&24&12&30\\Profit&2&4&10\\Profit/Steel&.2&.2&.4\\Profit/Al&.083&.333&.333\end {array}$$  With this data, the $10$ speed bikes are more profitable than either of the others per unit of each material, so we make as many of those as possible.  Based on steel, we can make $\frac {50075}{25}=2003$ and based on aluminum, we can make $\frac {62160}{30}=2072$.  If we make all those $10$ speeds, we use $60090$ aluminum and have $2070$ left.  Since neither of the other makes more profit per aluminum than $10$ speeds, it doesn't make sense to trade off some steel and make another model.
