Question 1210328
<pre>
Let x = the number of school uniforms
Let y = the number of social clothes

Maximize {{{C=600x+360y}}}
subject to the constraints
{{{x>=3}}}
{{{y>=3}}}
{{{600x+360y<=3600}}}

Simplify the third constraint by dividing through by 120

Maximize {{{C=600x+360y}}}
subject to the constraints
{{{x>=3}}}
{{{y>=3}}}
{{{5x+3y<=30}}}

In a first quadrant graph, we graph the constraint lines:
x=3, y=3, 5x+3y=30 

{{{drawing(400,23600/39,-.9,6.9,-.9,10.9,
graph(400,23600/39,-.9,6.9,-.9,10.9),

line(0,3,7,3),line(3,0,3,11), line(0,10,6,0),
locate(3.1,5.1,"(3,5)"),
locate(4.2,3.4,"(4.2,3)"),
locate(2.2,3.4,"(3,3)"),
locate(3.15,3.8,"F.R."),
locate(3.6,3,"(4,3)"),
locate(2.2,4.2,"(4,3)"),
circle(4,3,.07), circle(4,3,.04),circle(4,3,.035),circle(4,3,.02),
circle(3,3,.07), circle(3,3,.04),circle(3,3,.035),circle(4,3,.02),
circle(3,5,.07), circle(3,5,.04),circle(3,5,.035),circle(3,5,.02),
circle(3,4,.07), circle(3,4,.04),circle(3,4,.035),circle(3,4,.02)
 )}}}

The feasible region is 
on and above the line y=3
on and right of the line x=3
on and below the line 5x+3y=30
The feasible region is the triangle marked F.R., but since values must be
integers, only the lattice points are feasible. These are all marked.
The corner points of the feasible region are found by solving the systems
{{{system(x=3,y=3)}}}, {{{system(y=3,5x+3y=30)}}}, {{{system(x=3,5x+3y=30)}}}
The corner points are (3,3), (4.2,3), and (3,5)

But the corner point (4.2,3) is not a feasible point because we can
only choose lattice points (which have both coordinates as non-negative
integers) in the feasible region. So we choose the corner point
(4,3), as the feasible point nearest the corner point (4.2,3).

corner point of   
feasible region        
(or feasible point
nearest corner point)    Value of C

   (3,3)              600(3)+360(3) = 2880 
   (4,3)              600(4)+360(3) = 3480
   (3,5)              600(3)+360(5) = 3600

a) in how many ways can he spend the money?

This includes every feasible point within the entire feasible region.
These are (3,3),(3,4), (3,5), (4,3)
So the answer is there are 4 ways to spend the money

b) Which of these ways uses the MOST (not "much") of the N3600

The feasible point (3,5) has the maximum money spent at N3600, so this
is when he buys 3 uniforms and 5 social clothes.

Edwin</pre>