A Company has introduced a new product whose annual demand will depend on price charged. The demand of the product is represented by the function q = 100,000 – 200p, (where p = price and q = quantity demanded annually). Studies indicate that total cost of producing q unit is represented by TC = 150,000 + 100q + 0.003 q2 quantity to be produced to maximize profit is:

Tc=150000+100q+0.003q^2

Q=100000_200p
Max profit=?
At max profit ;MC=MR...eq (1)
Q=100000-200P
P=q_100000/_200
P=1/_200 q + 500
P=500-0.005q
TR=(500-0.005q)q
=500q-0.005q^2
MR=500-0.01q
From TC ; MC=100+0.006q
Putting MC and MR in eq (1)
500_0.01q=100+0.006q
500_100=0.006q+o.o1q
400=0.016q
q= 25000....ANSWER ..CHEERS

To maximize profit, we need to determine the quantity that will yield the highest difference between total revenue and total cost. This can be done by finding the quantity at which the derivative of the profit function is equal to zero.

The profit function can be calculated as follows:
Profit = Total Revenue - Total Cost

Total Revenue = Price * Quantity
Total Cost = 150,000 + 100q + 0.003q^2

Substituting the given demand function (q = 100,000 - 200p) into the total revenue equation, we get:
Total Revenue = p * (100,000 - 200p)

Now, we can write the profit function as follows:
Profit = (p * (100,000 - 200p)) - (150,000 + 100q + 0.003q^2)

Expanding and simplifying the profit function, we have:
Profit = 100,000p - 200p^2 - 150,000 - 100q -0.003q^2

To find the quantity that maximizes profit, we need to find the derivative of the profit function with respect to quantity (q) and set it equal to zero:

Profit' = 0 - 100 - 0.006q

Setting Profit' equal to zero:

-100 - 0.006q = 0
-0.006q = 100
q = 100/(-0.006)
q ≈ 16,667

Hence, the quantity that should be produced to maximize profit is approximately 16,667 units.

To find the quantity that maximizes profit, we need to find the quantity where revenue (R) minus cost (C) is maximized. Profit (P) is calculated by subtracting the cost from the revenue, so we have the equation:

P = R - C

First, let's find the revenue (R). Revenue is the product of the price (p) and the quantity demanded (q), so:

R = p * q

Substituting the given demand function, we have:

R = p * (100,000 - 200p)
R = 100,000p - 200p^2

Next, let's find the cost (C). The given cost function is:

TC = 150,000 + 100q + 0.003q^2

Substituting q = quantity demanded, we have:

C = 150,000 + 100q + 0.003q^2

Now, we can substitute the revenue (R) and cost (C) into the profit equation:

P = R - C
P = (100,000p - 200p^2) - (150,000 + 100q + 0.003q^2)

Simplifying the equation, we have:

P = 100,000p - 200p^2 - 150,000 - 100q - 0.003q^2

To maximize profit, we need to find the quantity (q) that maximizes this equation. To do that, we can take the derivative of the profit equation with respect to the quantity (q), and set it equal to zero:

dP/dq = 0

Differentiating the profit equation with respect to q, we have:

dP/dq = -200 - 0.006q - 100

Setting this equal to zero:

-200 - 0.006q - 100 = 0

Simplifying, we have:

-0.006q = 300

Dividing both sides by -0.006:

q = 300 / -0.006
q = 50,000

Therefore, the quantity that should be produced to maximize profit is 50,000 units.