Raggs, Ltd. a clothing firm, determines that in order to sell x suits, the price per suit must be pequals160 minus 0.75 x. It also determines that the total cost of producing x suits is given by Upper C (x )equals4000 plus 0.5 x squared. a) Find the total revenue, Upper R (x ). b) Find the total profit, Upper P (x ). c) How many suits must the company produce and sell in order to maximize profit? d) What is the maximum profit? e) What price per suit must be charged in order to maximize profit?

a) R(x) = 160x - 0.5x^2

b) P(x) = 160x - x^2 - 4000

c) The company must produce and sell 107 suits in order to maximize profit.

d) The maximum profit is $1,671.

e) The price per suit that must be charged in order to maximize profit is $106.50.

Explanation:

Given;

Price = p = 160 - 0.5x , (1)

Total cost = C(x) = 4000 + 0.5x^2 (2)

We can solve as follows:

a) Find the total revenue

Total revenue = R(x) = p * x (3)

Since from equation (1) p = 160 - 0.75x, we therefore substitute into equation (3) solve to have:

R(x) = (160 - 0.5x)x

R(x) = 160x - 0.5x^2 (4) < Total revenue

b) Find the total profit, Upper P(x).

P(x) = R(x) - C(x) (5)

Substituting equations (2) and (4) into equation (5) and solve, we have:

P(x) = 160x - 0.5x^2 - (4000 + 0.5x^2)

P(x) = 160x - 0.5x^2 - 4000 - 0.5x^2

P(x) = 160x - x^2 - 4000 (6) < Total profit

c) How many suits must the company produce and sell in order to maximize profit?

Profit is maximized where Marginal Revenue (MR) is equal to Marginal Cost (MC). That is where;

MR = MC (7)

Where MR = price = p = 160 - 0.5x

MC is obtained buy differentiating equation (2) with respect to x as follows:

MC = C'(x) = x

Substituting for MR and MC in equation (7) and solve for x, we have:

160 - 0.5x = x

160 = x + 0.5X

160 = 1.5x

x = 160 / 1.5

x = 107

Therefore, the company must produce and sell 107 suits in order to maximize profit.

d) What is the maximum profit?

To obtain this, we substitute x = 107 into equation (6) and solve as follows:

P(x) = 160(107) - 107^2 - 4000

P(x) = (160 * 107) - 107^2 - 4000

P(x) = 17,120 - 11,449 - 4000

P(x) = 1,671

Therefore, the maximum profit is $1,671.

e) What price per suit must be charged in order to maximize profit?

To obtain this, we substitute x = 107 into equation (1) and solve as follows:

p = 160 - 0.5(107)

p = 160 - (0.5 * 107)

p = 160 - 53.50

p = 106.50

Therefore, price per suit that must be charged in order to maximize profit is $106.50.

answer; /// the use of a bank credit card to make purchase;

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