Explain every part All created functions mu st have units Se

Explain every part. All created functions mu st have units. See note at end before starting. An exclusive health club has fixed costs of $11,000 per month and a variable cost of $35 per month per health club member. Find the cost function (dollars spent per month by the club) as a function of the number of club members. The demand for membership is such that when the demand is 1000 members then the price is $100 per month, (in other words. 1000 people would join the health club if the cost was $100 per month). When the demand is 1800 members then the price is $60 per month. Assume that demand is a linear function. Find the demand function (monthly price per member) as a function of the number of club members. At what monthly price per member will there be no members? Find the revenue function (monthly income in dollars) as a function of the number of club members Write a profit function (monthly dollars of profit) as a function of the number of club members. What will be the monthly profit (or loss) for having no members? For what number of club members will the club break even? For what monthly prices does the club make a profit? Calculate the monthly membership that will maximize profit for the club. What would that profit be? What will be the monthly price per member if maxi mu m profit is reached?

Solution

a) Cost = fixed cost + variable cost

Let m be the number of members

Fixed costs = 11,000 Variable costs = 35 per member = 35 * m

Hence cost function C(m) = 11000+35m ...............(1)

b)Since the demand function is linear, we know that it can be written in slope-intercept form as

p = am+b where p is the price per member

When m=1000, p = 100: 100 = 1000a + b ..........(i)

When m =1800, p = 60: 60 = 1800a + b ..............(ii)

Subtracting equation (ii) from (i) we get, 40= -800a this gives a = -40/800= -0.05

Substituting a = -0.05 in equation (i) we get 100 = 1000(-0.05)+b which implies 100= -50 +b

Therefore b = 100+50 = 150. Hence the demand function is p = -0.05m + 150........(2)

c) we need to find p for m = 0. Substitute m =0 in eqn(2)

p = 150 Hence there will be no members for monthly price 150 per member

d) The revenue function is found by observing that revenue = price x quantity

Since we are looking for revenue as a function of the quantity m,we’ll rewrite this as R(m)= pm

R(m) = (-0.05m+150)m = -0.05m²+150m

e) profit = Revenue-cost

P(m) = R(m) - C(m)= -0.05m²+150m - (11000+35m)

P(m)= -0.05m²+115m - 11000

f) For no members m = 0

P(0) = -11000

Hence for no members there will be a loss of $11,000

g) for break even point C(m) = R(m)

11000 + 35m = -0.05m²+150m

-0.05m²+115m - 11000=0

Dividing through out by -0.05 we get

m² - 2300m + 220000 = 0

m²- 2200m - 100m + 220000= 0

Factoring we get

(m-2200)(m-100) = 0

Which means m=100 or m= 2200

So the club will break even for the first 100 members.

h) for profit we need to solve

P(m) > 0 which implies -0.05m²+115m - 11000 > 0 i.e. 100<m<2200

P(100) = -0.05(100)+150 = 145; P(2200)= -0.05(2200)+150= 40

Hence for Profits price per member is between 40 and 145

i) For maximum profit we look at the axis of symmetry of the above quadratic equation

m =-b/2a = -115/2(-0.05)=1150. So 1150 members will maximize profit

k) Cost per member for maximum profit= -0.05(1150)+150 = $92.5

j) P(1150) = 55,125 Hence max profit = $55,125