Question 1210328: A student requires at least three school uniforms and three social clothes. The uniform costs N600 and the social wear costs N360 each. If the student has N3 600 to spend and he decides to spend as much as possible of the N3 600, a) in how many ways can he spend the money? b) Which of these ways uses the much
(MOST, not "much") of the N3600
Found 4 solutions by mccravyedwin, greenestamps, Edwin McCravy, ikleyn:
Answer by mccravyedwin(408)   (Show Source):
You can put this solution on YOUR website!
Let x = the number of school uniforms
Let y = the number of social clothes
Maximize 
subject to the constraints
Simplify the third constraint by dividing through by 120
Maximize
subject to the constraints
In a first quadrant graph, we graph the constraint lines:
x=3, y=3, 5x+3y=30
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
 ,  , 
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
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
The other tutor has provided a response showing a very formal mathematical solution.
Here is an informal solution.
Start by buying the minimum required number of uniforms and social clothes. The cost is 3(360) + 3(600) = 1080+1800 = 2880.
The amount left to spend is 3600-2880 = 720.
You can of course buy only the required minimum number of items, so the first possible way to spend the money is 3 uniforms and 3 social clothes.
(1) (3,3); 2880
With 720 remaining, you can only buy one more uniform at 600 each. So a second possibility is 4 uniforms and 3 social clothes.
(2) (4,3); 3480
Or, with the remaining 720, you can buy either 1 or 2 more sets of social clothes at 360 each.
(3) (3,4); 3240
(4) (3,5); 3600
ANSWERS:
(a) There are 4 ways to spend the money
(b) Buying 3 uniforms and 5 sets of social clothes uses the largest amount (all) of the available money
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
Indeed, since this problem has only 4 possible ways to buy the uniforms and
social clothes and spend no more than N3600, it can be solved informally.
However, I believe this problem was assigned to lead the student to understand
linear programming problems, which require graphical solutions, for instance,
like this one:
A factory manufactures two types of gadgets, regular and premium. Each gadget
requires the use of two operations, assembly and finishing, and there are at
most 12 hours available for each operation. A regular gadget requires 1 hour of
assembly and 2 hours of finishing, while a premium gadget needs 2 hours of
assembly and 1 hour of finishing. Due to other restrictions, the company can
make at most 7 gadgets a day. If a profit of $20 is realized for each regular
gadget and $30 for a premium gadget, how many of each should be manufactured to
maximize profit?
This problem requires graphical methods, and no doubt is the sort of problem
this student will very soon be required to solve. That's why I chose to solve
the given problem as close to the way to solve the above problem as possible.
Edwin
Answer by ikleyn(52814)  (Show Source):
You can put this solution on YOUR website! .
A student requires at least three school uniforms and three social clothes.
The uniform costs N600 and the social wear costs N360 each.
If the student has N3 600 to spend and he decides to spend as much as possible of the N3 600,
a) in how many ways can he spend the money?
b) Which of these ways uses the much
(MOST, not "much") of the N3600
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In this problem, the number of points under the interest is 4.
Imagine for a minute, that there are 40 or 70 solutions in integer numbers
in such or similar problem. How will you find them ?
There is very mechanical way to do it, practically without thinking.
Plot the restriction lines in a graph paper.
Then you will see the integer grid points inside the feasible area
(may be, including those on the boundaries).
Count all of them, and you will get the answer.
Such plot can be produced electronically, using appropriate plotting tool, for example, in this site
https:\\www.desmos.com/calculator/
You only need to take care about units on the axes, in order for
the points of interest were all integer points of the grid.
For it, you should simplify your restriction equations to make them
as simple as possible - - - or to select the units in the axes by an appropriate way.
I do not state that this method will work everywhere or always, but in many practical cases
(in 95% typical school Math problems of this kind) it will work effectively.
So, keep it in your mind.
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