Math - Nonlinear Models?

Question

Television Advertising

As sales manager for Monevideo Production, Inc, You are planning to review the price you charge clients fot TV advertisement development.You currently change each client an hourly development fee of $2500. With this pricing strcuture, the demand, measured by the number of contracts Montevideo sings per month, is 15 contracts. This is down 5 contracts from the figure last year, when your company charged only $2000.

a. Construct a linear demand equation giving the number of contracts q asa function of the hourly fee p Montevideo charges for development.

b. On average, Montevideo bills for 50 hours of production time on each contract. Give a formula for the total revenue obtained by charging $p pre hour.

c. the costs to Montevideo Productions are estimated as follows:


Fixed costs: $120,000 per month
Variable costs: $80,000 per contract

Express Montevidso Productions'monthly cost (i) as a functionof the number q of contractss and (ii) as a function of the hourly production charge p.

d. Express Monteviedo Productions' monthly profit as a function of the hourly development fee p and find the price it should charge to miximize the profit.

Answer 1

I like keeping units around. Makes it look worse, but makes sure that you're not adding apples and oranges.

a) You are given two points: ($2,500, 15 contracts per month) and ($2,000, 20 contracts per month).

You can use those to find the slope of the line:

m = (15 contracts per month - 20 contracts per month)/($2,500 - $2,000) = -5 contracts per month/$500 = -1 contract per month/$100

(or m = (15 - 20)/(2,500 - 2,200) = 0.01)

which means that every $100 you raise the price, you lose a contract.

Then use either point to find an equation:

(q - 15 contracts per month) = -1 contract per month/$100 (p - $2,000)
q - 15 contracts per month = -1 contract per month/$100 × p + 20 contracts per month
q(p) = -1 contract per month/$100 × p + 35 contracts per month

(or, if you don't like seing the units: q(p) = -0.01p + 35)

b) This is 50 hours per contract times $p per hour times the total number of contracts from part a:

(-1 contract per month/$100 × p + 35 contracts per month) × 50 hours/contract per month × $p/hour =
TR(p) = -$(1/2 × p²) + $(1,750 × p)

(or without units, TR(p) = -1/2 p² + 1,750 p)

c) Total cost = fixed cost + variable cost per contract × number of contracts:

i) TC(q) = $120,000 per month + $80,000/contract × q contracts per month

(or TC(q) = 120,000 + 80,000q)

ii) As a function of p, you can use the formula above for q in terms of p:

TC(p) = $120,000 per month + $80,000/contract × (-1 contract per month/$100 × p + 35 contracts per month)

or

TC(p) = $120,000 per month - $(800 × p) per month + $2,800,000 per month

or

TC(p) = $2,920,000 per month - $(800 × p) per month

(TC(p) = 2,920,000 - 800p)

d) OK, I liked the units, but I'm gonna stop now.

So far we have, if I haven't messed up:

TR(p) = -1/2 p² + 1,750p
TC(p) = 2,920,000 - 800p

And since total profit per month = total revenue per month - total cost per month:

TP(p) = TR(p) - TC(p)
TP(p) = -1/2 p² + 1,750p - (2,920,000 - 800p)
TP(p) = -1/2 p² + 2,550p - 2,920,000

There are at least two ways to find the best price. Hopefully you'll understand at least one of them.

Derivative method (calculus):

Take the derivative and set it equal to zero:

TP'(p) = -p + 2550 = 0, so p = $2,550 per contract will maximize the monthly profit.

TP(2550) = -1/2 (2,550)² + 2,550(2,550) - 2,920,000 = $331,250 per month

Parabola method (algebra 2):

TP(p) = -1/2 p² + 2,550p - 2,920,000

TP(p) = -1/2(p² - 5,100p) - 2,920,000

TP(p) = -1/2(p² - 5,100p + 6,502,500) - 2,920,000 + 3,251,250

TP(p) = -1/2(p - 2,550)² + 331,250

So the maximum of the function is the point ($2550 per contract, $331,250 per month)