A calculator company plans to sell two models of graphing calculators that cost $100 and $150, respectively. The $100 model yields a profit of $40 and the $150 model yields a profit of $50. The company estimates that the total monthly demand will not exceed 250 units. What are the number of units of each model should be stocked in in order to maximize profit, assuming that the merchant does not want to invest more than $30,000 in inventory?

The quantity of the first model is 150 and the second model is 100 that maximize the profit.

Step-by-step explanation:

Let the quantity of first model = x

Let the quantity of second model = y

The cost of the first model = $100

The cost of the second model = $150

Total number of models = 250 units

Total amount to spend on units = $30000

Now form the equations.

x  +  y  = 250

100x + 150y = 30000

Now solve for the x and y.

x = 250 – y

Now insert this value in the 100x + 150y = 30000.

100(250 –y) + 150y = 30000

25000 – 100y + 150y = 30000

50y = 30000 – 25000

50y = 5000

Y = 100

Now insert the Y in x + y = 250

x = 250 – y

x = 250 – 100

x = 150

Therefore, the first model is 150 units and the second model is 100 units that maximize the profit.

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

answer: what is the question you're trying to ask?

step-by-step explanation: