For xyz manufacturing the fixed costs are $1200, material and labor costs combined are $2 per unit, and the demand equation is:

Question

For xyz manufacturing the fixed costs are $1200, material and labor costs combined are $2 per unit, and the demand equation is:

p=100/√q

What level of output will maximize profit? Show this occurs when marginal revenue equals marginal cost. What is the price at the profit maximization?

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I understand R=p*q

R=100/√q * q =100/√q

I think
Profit = R-2q-1200

Then I don't know what to do.. Can anyone help?

Answer 1

You have a slight(!) error:

R = 100/√q * q = 100√q

Profit = R-2q-1200
= 100√q - 2q - 1200

maximum profit where dP/dq = 0

50/√q - 2 = 0
50 = 2√q
q = 625

since P = R-C,
dP/dq = 0 when dR/dq = dC/dq

The price is 100*25 = 2500

Answer 2

To find the level of output that maximizes profit, we need to find when marginal revenue equals marginal cost. Let's start by finding the expressions for marginal revenue and marginal cost.

The marginal revenue (MR) is the derivative of the revenue equation with respect to the quantity (q). In this case, the revenue equation is R = p*q, where p is the price and q is the quantity. Since p = 100/√q, we can substitute this expression into the revenue equation:

R = (100/√q) * q = 100√q

Now, to find the marginal revenue, we take the derivative of R with respect to q:

MR = dR/dq = 100(1/2)(q^(-1/2)) = 50/q^(1/2)

Next, let's find the expression for marginal cost (MC). The fixed costs are $1200, and the material and labor costs combined are $2 per unit, which means the variable cost per unit (VC) is $2. Therefore, we can express the marginal cost as:

MC = d(VC*q)/dq = 2

Now, we can set MR equal to MC and solve for the quantity that maximizes profit:

MR = MC
50/q^(1/2) = 2

To simplify the equation, let's square both sides:

(50/q^(1/2))^2 = 2^2
2500/q = 4

Now, let's solve for q:

q = 2500/4
q = 625

The level of output that maximizes profit is 625 units.

To find the price at the profit maximization, we can substitute the value of q into the demand equation p = 100/√q:

p = 100/√(625)
p = 100/25
p = 4

Therefore, the price at the profit maximization is $4.

Answer 3

To find the level of output that maximizes profit, we need to determine the quantity at which marginal revenue (MR) equals marginal cost (MC).

First, let's find the marginal revenue (MR) by taking the derivative of the demand equation with respect to q:

MR = d(R)/dq = d(100/√q)/dq

To simplify this, let's rewrite the demand equation as:
R = 100q^(-1/2)

Taking the derivative, we have:
MR = d(100q^(-1/2))/dq = -50q^(-3/2)

Next, let's find the marginal cost (MC):
MC = d(C)/dq = d(2q + 1200)/dq = 2

For profit maximization, MR = MC. Therefore, we can set -50q^(-3/2) = 2:

-50q^(-3/2) = 2

To solve for q, let's invert both sides of the equation:

-1/(50q^(-3/2)) = 1/2

Now, solving for q, we get:

q^(-3/2) = 1/100

Taking the reciprocal of both sides:

q^(3/2) = 100

Taking the square root of both sides:

q = (100)^(2/3) = 31.62 (approximately)

So, the level of output that maximizes profit is approximately 31.62 units.

To find the price at the profit maximization, substitute this quantity value back into the demand equation:

p = 100/√q
p = 100/√31.62
p ≈ 100/5.63
p ≈ 17.75

Therefore, the price at the profit maximization is approximately $17.75 per unit.