Question

Raggs, Ltd. a clothing firm, determines that in order to sell x suits, the price per suit must be p=140-0.5x. It also determines that the total cost of producing x suits is given by C(x) = 2000 +0.75x^2

a) Find the total revenue, R(x)
b) Find the total profit, 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?

Answer 1

Answer:

(a) The total revenue is $R(x)=140x-0.5x^2$

(b) The total profit is $P(x)=-1.25x^2+140x-2000$

(c) The profit is maximized when 56 suits are produced and sold.

(d) The maximum profit is $1920

(e) The price per unit needed to make the maximum profit is $p=\ 112$

Step-by-step explanation:

From the information given we know:

• The price per suit must be $p=140-0.5x$
• The total cost of producing x suits is given by $C(x) = 2000 +0.75x^2$

(a) The total revenue is R(x) = (Numbers of units) x (Price per unit)

$R(x)=x\cdot (140-0.5x)\\\\R(x)=140x-0.5x^2$

(b) The total profit is P(x) = Total revenue - Total cost

$P(x)=R(x)-C(x)\\\\P(x)=(140x-0.5x^2)-(2000 +0.75x^2)\\\\P(x) = 140x-0.5x^2-2000-0.75x^2\\\\P(x)=-1.25x^2+140x-2000$

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

$\frac{d}{dx}P(x)=\frac{d}{dx}(-1.25x^2+140x-2000)\\ \\P'(x)=-\frac{d}{dx}\left(1.25x^2\right)+\frac{d}{dx}\left(140x\right)-\frac{d}{dx}\left(2000\right)\\\\P'(x)=-2.5x+140$

Next, we need to find all critical points P'(x) = 0

$-2.5x+140=0\\-2.5x\cdot \:10+140\cdot \:10=0\cdot \:10\\-25x+1400=0\\-25x+1400-1400=0-1400\\-25x=-1400\\\frac{-25x}{-25}=\frac{-1400}{-25}\\x=56$

Use the second derivative test to determine whether we have an absolute maximum because f(c) is the absolute maximum value if f''(c) < 0.

$\frac{d}{dx} P'(x)=\frac{d}{dx}(-2.5x+140)\\\\P''(x)=-\frac{d}{dx}\left(2.5x\right)+\frac{d}{dx}\left(140\right)\\\\P''(x)=-2.5$

Thus, P''(56) is negative, and so profit is maximized when 56 suits are

produced and sold.

(d) The maximum profit is given by

$P(56)=-1.25(56)^2+140(56)-2000\\\\P(56)=-56^2\cdot \:1.25+7840-2000\\\\P(56)=5840-56^2\cdot \:1.25\\\\P(56)=5840-3920\\\\P(56)=\ 1920$

The clothing firm makes a maximum profit of $1920 by producing

and selling 56 suits.

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

$p=140-0.5(56)=\ 112$

Answer 2

(a) The total revenue is R(x) = 140x - 0.5x^2

(b) The total profit is P(x)= -1.25x^2 +140x - 2000

(c) The profit is maximized when 56 suits are produced and sold.

(d) The maximum profit is $1920

(e) The price per unit needed to make the maximum profit is p= $112

Calculations and Parameters:

Given that

• The price per suit must be p= 140- 0.5x
• The total cost of producing x suits is given by c(x) = 2000 + 0.75x^2

The total revenue is R(x) = (Numbers of units) x (Price per unit)

The total profit is P(x) = Total revenue - Total cost

To find the maximum value of P(x) we need to find the derivative P'(x) which gives us P"(x)= -2.5.

Thus, P''(56) is negative, and so profit is maximized when 56 suits are produced and sold.

The maximum profit is given by P(56)= 5840 - 3920

=$1,920

The price per unit needed to make the maximum profit is P= 140- 0.5(5.6) = $112.

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