Probem57+Max+Ian+Carolyn

a) Carolyn: Linear equation: y=mx + b m= (1100-1000)/(440-450) m= -10 D(x)=-10x + b 1000= -10(450) + B 1000+4500= b b= 5500

D(x)=-10x + 5500

R(x)= xD(x) R(x)=x(-10x +5500)

b) Maximize R(x) R(x)= -10x^2 + 5500x R'(x) = -20x + 5500 0= -20x + 5500 -5500/-20=x x= 275

So if they charge $275 per TV, then they should offer a rebate of $175 ^Do we not need to check endpoints.. what does the endpoint behavior suggest in this problem? yes you do. They could certainly sell televisions at $0 (and the demand would be high!)...but what is the maximum value at which they could sell television? In other words, (and looking at your demand function) what is the price of the television for which they would sell 0 sets?

c) P(x)= R(x)-C(x) in this situation C(x) refers to the cost of producing x televisions, but your Revenue function does not involve x televisions! P(x)= -10x^2 + 5500x - (68000 + 150x) P(x)= -10x^2 + 5500x -68000 -150x P(x)= -10x^2 + 5350x -68000 P'(x)= -20x + 5350 0=-20x + 5350 x=5350/20 x=267.5 ^I tried this with R(D(x)) and I could not get a real solution. But didn't we already take D(x) into account in part a when we defined R(x) as xD(x) ? Therefore they should offer a rebate of $182.50

Demand = # of tv's sold

sets/week = 1000 + 100x, where x is equal to the number of rebates offered

c) Using your work above, use the fact that Profit is total Revenue minus total Cost. The function C(x) that you are given is the cost of producing x items....but our x above is the price we have set. If x is our price, what did we determine the number of items we will sell will be?

Ian R. Wilson