Question

Silver Scooter Inc. finds that it costs $100 to produce each motorized scooter and that the fixed costs are $1,000. The price is given by p = 700 - X, where p is the price in dollars at which exactly x scooters will be sold. Find the quantity of scooters that the company should produce and the price it should charge to maximize profit. Find the maximum profit How many scooters should the company produce to maximize profit? scooters What price should the company charge to maximize profit? What is the maximum profit?

Answer 1

Solution-

(1)

Since fixed cost of producing the scooter is $ 1000 and variable cost is $ 100 per scotters

So, cost price in dollars for producing x scooters is

CP = 1000 + 100x ....(1)

Since selling price of one scooter is p = 700 - x.

So, selling price of x scooters is

SP = p×x

SP = (700 -x)(x)

SP = 700x - x² .....(2)

From equation (1) and (2) , we get

The profit on selling x scooters in dollars is

P(x) = SP -CP

P(x) = (700x - x² ) - (1000 +100x)

P(x) = 700x - x² - 1000 - 100x

P(x) = - x^{2​​​​​​} + 600x - 1000 ....(3)

Now,

(a)

To have maximum profit, the differentiation of profit function must be equal to 0.

Differentiating equation (3) and putting it equal to 0.

So, P'(x) =0 implies

(d/dx)(- x² +600x +1000) =0

Or -2x +600(1) + 0 =0

Or -2x = -600

Or x = -600/(-2)

Or x = 300

Hence, for maximum profit, number of scooters the company have to produce are x = 300.

(b)

Putting x = 300 in price equation we get

P = 700 - x = 700 - 300 = 400 dollars

Hence, to maximize the profit, company should charge $ 400 per scooter.

(c)

Putting x =300 in equation (3) to get

P(300) = - (300)² +600(300) - 1000

P(300) = -90000 + 180000 - 1000

P(300) = $ 89,000

Hence, the maximum profit is $ 89,000 .

(2)

Let pressure, temperature and volume are denoted by P, T and V respectively.

Now, according to question

Pressure of a gas in a container is directly proportional to temperature and inversely proportional to volume.

So, it can be written as

P = k(T/V) ........(1)

Here k is constant.

Since at Temperature T = 70 Kelvin , Volume V =15 m³ the pressure is P = 16.8 N/m² .

On putting these values in equation (1), we get

16.8 = k(70/15)

Or k(70/15) = 16.8

Or k = 16.8×15/70

Or k = 3.6

Putting k =3.6 in equation (1), we get

P = 3.6(T/V)

Now, at T = 80 Kelvin, V = 8 m³

Pressure P = 3.6(T/V) = 3.6(80/8) = 36 N/m²

Hence, Pressure is 36 Newtons per square meters.

(3)

Consider the demand equation for p price and x television sold is

p = -0.17x + 187 .....(1)

For selling selling television, the revenue function is

R(x) = p×x

R(x) = (-0.17x + 187)(x)

R(x) = -0.17x² + 187x .....(2)

Now,

(a)

Putting x = 70 units ( television in stock) in equation (1), we get

p = -0.17(70) + 187

p = -11.9 + 187 =$ 175.10

Hence, the charge for each unit in stock is $ 175.10 .

(b)

To maximize the revenue function, put differentiation of equation (2) equal to 0.

So, R'(x) =0 implies

(d/dx)(-0.17x² + 187x) =0

Or -0.17(2x) + 187(1) = 0

Or -0.34x + 187 =0

Or -0.34x = -187

Or x = 187/(-0.34)

Or x = 550 units

x = 550 units will maximize the revenue.

(c)

Putting x = 550 units in equation (2) to get

Maximum Revenue

R(550) = -0.17(550)² + 187(550)

R(550) = - 51,425 + 102,850

R(550) = $ 51,425

Hence, maximum revenue is $ 51,425 .

(d)

On putting x = 550 units in equation (1) we get

Price at maximum revenue

p = -0.17(550) + 187

p = -93.5 + 187 =$ 93.5

Hence, price per television for maximum revenue is $ 93.5 .