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Question 1194100: A charity worker is going to sell wigs, candles and hat clips to raise funds for their organization. The cost a wig for $10 and sell it at the price of $30, the cost of a candle is $ 4 and they sell them for $ 12 and the hat clips cost $ 10 and sell them for $24. They have the maximum budget to buy the stuff is $ 4000. the area of a table is 60" long and 30" wide. The area occupied by the wigs is 10square inch, the candle occupied 6square inch and the hat clips 2 square inch. How should the charity organize their stuff so they can get the maximum profit?
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52787)   (Show Source):
You can put this solution on YOUR website! .
My previous solution in this post was incorrect, so I deleted it.
The logic I used was defective - - - thanks to Edwin, who detected and noticed it.
Use Edwin' solution, which is correct.
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
Here is the step-by-step solution using the standard simplex method:
Suppose they will buy w wigs, c candles, and h hats, for a maximum profit P.
[I used "hats" instead of "hat clips" for convenience]
|wigs |candles| hats|
------------------------------|-----|-------|-----|------------
How many bought and sold | w | c | h |
Cost each | 10 | 4 | 10 |
Cost for all bought & sold. |$10w | $4c |$10h | ≤ $4000 <--cost
Selling price for each |$30 |$12 |$24 |
Total money taken in |$30w |$12c |$24h |
Profit for all bought & sold |$20w | $8c |$14h | = P <--profit
Sq. in. required | 10w | 6c | 2h | ≤ 1800 <--sq. in.
So the problem is:
Maximize P = 20w+8c+14h <--objective function
subject to the constraints:
10w + 4c + 10h ≤ 4000
10w + 6c + 2h ≤ 1800
w ≥ 0, c ≥ 0, h ≥ 0
Rewrite the objective function as
-20w - 8c - 14h + P = 0
Introduce slack variables s1 and s2 to change the inequalities
into equations, and put the rewritten objective function at
the bottom:
Next, we put the above system in a special 3x7 partitioned
matrix called a "tableau", with each variable written
above its column of coefficients. The constants in the
right-most column are headed k:
| w c h |s1 s2 | P | k |
-------------------------------
| 10 4 10 | 1 0 | 0 | 4000 |
| 10 6 2 | 0 1 | 0 | 1800 |
-------------------------------
|-20 -8 -14 | 0 0 | 1 | 0 |
We find the most negative number ('indicator') on the
bottom row, which is -20. Its column, the 1st column,
is the "pivot" column.
We divide each POSITIVE number in the k-column by the
POSITIVE number in the pivot column which is in the
same row. That is:
4000/10=400, 1800/10=180
Since 180 is smaller, the pivot row is the 2nd row.
The pivot element is the 10 in the 2nd row.
We use row operations to make the pivot element become
1 and all the other elements in the pivot column become 0.
Here are three row operations we perform on the above
tableau matrix:
(1/10)R2 -> R2
-10R2+R1 -> R1
20R2+R3 -> R3
| w c h |s1 s2 | P | k |
-------------------------------
| 0 -2 8 | 1 -1 | 0 | 2200 |
| 1 .6 .2 | 0 .1 | 0 | 180 |
-------------------------------
| 0 4 -10 | 0 2 | 1 | 3600 |
We still have a negative number on the bottom row, -10.
Now we find the most negative number ('indicator') on the
bottom row. It happens to be the ONLY negative number on
the bottom row. which is -10. Its column, the 3rd column,
is the "pivot" column.
We divide each POSITIVE number in the k-column by each
POSITIVE number in the pivot column which is in the
same row:
2200/8=400, 180/.2=1800
Since 400 is smaller, the pivot row is the 1st row.
The pivot element is the 8 in the 1st row.
We use row operations to make the pivot element become
1 and all the other elements in the pivot column become 0.
Here are the row operations we perform on the above:
(1/8)R1 -> R1
-.2R1+R2 -> R2
10R1+R3 -> R3
| w c h | s1 s2 | P | k |
----------------------------------------
| 0 -.25 1 | .125 -.125 | 0 | 275 |
| 1 .65 0 |-.025 .125 | 0 | 125 |
----------------------------------------
| 0 1.5 0 | 1.25 .75 | 1 | 6350 |
This tableau is finished because there are
no more negative numbers (indicators) on
the bottom row.
We transform the tableau matrix back into
a system of equations:
That simplifies to:
Solve the bottom equation for P
Since c, s1 and s2 cannot be negative, we see that
the largest value P can take on is P = 6350 when
c = s1 = s2 = 0. So they will not buy any candles!!
So we substitute in the other two equations to find
the number of wigs and hats:
h = 275, w = 125
So the most lucrative plan is to buy
125 wigs, 0 candles, and 275 hats. and the maximum profit P = $6350
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