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?

Please refer to the below for answers.

Explanation:

As per the above information, the following can be deduced;

•The price per suit (P)= 160-0.75x

•Total cost of producing x suit is given as C(x) = 4,000+0.5x^2

(a) The total revenue is R(x);

= (Numbers of units) * (Price per unit)

R(x) = x * (160-0.75x)

R(x) = 160x - 0.75x^2

(b) The total profit is P(x);

= Total revenue - Total costs

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

P(x) = (160x - 0.75x^2) - (4,000 + 0.5x^2)

P(x) = 160x - 0.75x^2 - 4,000 - 0.5x^2

P(x) = -1.25x^2 + 160x - 4,000

(c). To find the maximum value of P(x), we will need to find first the derivative of P'(x)

d/dx*P(x) = d/dx(-1.25x^2 + 160x - 4,000)

P'(x) = - d/dx(1.25x^2) + d/dx(160x) - d/DX(4,000)

P'(x) = -2.5x + 160

The next is to find the critical points.

P'(x) = 0

-2.5x + 160 = 0

-2.5x . 10 + 160 . 10 = 0 . 10

-25x + 1,600 = 0

-25x + 1,600 - 1,600 = 0 - 1,600

-25x = -1,600

x = 64

We will also use the second derivative test to know if there is an absolute maximum because f(c) is the absolute maximum value if f''(c) < 0

Therefore,

d/dx*P'(x) = d/dx (-2.5x + 160)

P''(x) = -d/dx (-2.5x) + d/dx (160)

P''(x) = -2.5

Hence, (64) is negative , and so profit is maximized when 64 suits are produced and sold.

(d) The maximum profit is denoted as;

P(64) = -1.25(64)^2 + 160(64) - 4,000

P(64) = -64^2 * 1.25 + 10,240 - 4,000

P(64) = 6,240 - 64^2 * 1.25

P(64) = 6,240 - 5,120

P(64) = $1,120

Therefore , the clothing firms makes $1,120 as profit by producing and selling 64 suits.

(e) The price per unit to needed to make the maximum profit is thus;

P = 160 - 0.75x

Where x is 64

P = 160 - 0.75(64)

P = 160 - 48

P = $112

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