SOLUTION: 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 de

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(52818)   (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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