To sell 300 notebook and 120 pens quickly, a store decided to create two types of gift sets. One set is to include 2 notebooks and 1 pen, and the other will consist of 1 pen and 3 notebooks. First set will be sold for $8. While the second one is to cost $11.50. It is unprofitable for the store to sell less than 20 of any gift set. How many gift sets of each type does store need to sell to obtain the biggest possible income?
graph each of the lines of constraint, and shade the areas that satisfy the inequalities. The intersection of all those shaded area is your region of interest. The objective function p will achieve its maximum value at one of the vertices of the shaded region. So evaluate p at each vertex, plugging in the x,y values there. Pick the vertex where p is the largest. In this case, select the closest integer values, since pens and notebooks don't come in fractional amounts.
If this still makes no sense, review linear programming and maximizing objective functions. Many examples and explanations are easily found online.
If there are
x sets of 2 notebooks and 1 pen
y sets of 3 notebooks and 1 pen
Then you want to maximize p = 8x + 11.50y
subject to the constraints
2x+3y <= 200
x+y <= 120
x >= 20
y >= 20
So now just graph the region and check p at each vertex, as usual.
2x+3y <= 200
300 I think :(
the curse of the typo strikes again ...
good catch
But I'm sure Phoenix caught it too, right? ... right?
I have the same question, but I couldn't understand what you mentioned above oobleck
Got it, thanks :)
To determine the number of gift sets the store needs to sell to obtain the largest possible income, we can use a mathematical approach.
Let's assume the store sells x sets of the first type (2 notebooks and 1 pen) and y sets of the second type (1 pen and 3 notebooks).
We know that the store needs to sell at least 20 of each gift set. So, we have the following constraints:
x ≥ 20 (for the first type)
y ≥ 20 (for the second type)
The total number of notebooks in the first type of gift set would be 2x, and in the second type, it would be 3y. Similarly, the total number of pens in the first type of gift set would be x, and in the second type, it would be y.
Based on the given information, we can create the following equations:
2x ≤ 300 (total notebooks sold should be less than or equal to 300)
3y ≤ 120 (total notebooks sold should be less than or equal to 120)
x + y ≤ 120 (total pens sold should be less than or equal to 120)
Let's solve these equations:
From 2x ≤ 300, we find that x ≤ 150.
From 3y ≤ 120, we find that y ≤ 40.
From x + y ≤ 120, we find that x ≤ 120 - y.
Now, to maximize the income, we need to determine the number of gift sets that will give the highest total income.
The income from selling x sets of the first type would be 8x.
The income from selling y sets of the second type would be 11.50y.
So, the total income would be 8x + 11.50y.
As we want to maximize the income, we need to maximize the expression 8x + 11.50y, while considering the constraints.
Using these constraints, we can create a feasible region by plotting the inequalities on a graph and finding the area where they intersect.
After plotting the graph and finding the feasible region, we can see that the maximum income is obtained at the corner point where x = 150 and y = 40.
Therefore, the store needs to sell 150 sets of the first type (2 notebooks and 1 pen) and 40 sets of the second type (1 pen and 3 notebooks) to obtain the biggest possible income.