Question 42819
This is a LP (Linear Programming) problem. It is not an algebra problem at all. I have done such problems in our Industrial Management class last semester (also my final semester in Bachelor of Engineering). Though it is not algebra, I am solving this one. Here you go....


Let the no. of products A sold be 'x' and that of product B sold be 'y'.
Then cost of production of all the products = $(9x + 8y).
Selling price of all the products = $(12x + 10y).
Hence, overall profit = $(12x + 10y) - $(9x + 8y) = $(3x + 2y).
So the objective function is: f(x,y) = 3x + 2y.


The constraints are: 
{{{200 <= x <= 300}}}, {{{100 <= y <= 250}}} and {{{x + y <= 500}}}


{{{drawing(600,600,-600,600,-600,600,
line(-100,600,600,-100),
line(200,600,200,-600),
line(300,600,300,-600),
line(-600,100,600,100),
line(-600,250,600,250),
grid(1),
red(line(200,250,250,250), line(250,250,300,200), line(300,200,300,100), line(300,100,200,100), line(200,100,200,250)),
locate(200,250,A),
locate(250,250,B),
locate(300,200,C),
locate(300,100,D),
locate(200,100,E)
)}}}


Pentagon ABCDE is the required region bounded by constraints.


The vertices are A(200,250), B(250,250), C(300,200), D(300,100) and E(200,100).


For the vertex A(200,250), the value of objective function is f(x,y) = f(200,250) = {{{3*200 + 2*250}}} = 1100.
Similarly, the other objective functions at B, C, D and E are 1250, 1300, 1100 and 800 respectively.


Profit will be maximum when the number of units produced will be maximum.
Maximum number of units that can be procuded is 500 so the values of x & y for maximum profit will lie on line BC.
Keeping total number of units produced same, at 500 (which is maximum value), the profit is maximized when the no. of units of A [in which profit per unit ($3) is more than in B (2$)] produced is of the maximum possible value. This condition is satisfied by the coordinates of the point C.
Hence, for maximum profit 300 units of A and 200 units of B have to be produced and sold.


I have shown you the process.
Do the next one yourself because more you do these problems yourself, better you learn.