Hello, xcelxp,

to b) and c) only:

let h be the number how often the rebate was offered, then the revenue can be calculated by:

The graph of this function is a parabola which opens downward that means that the maximum is the r-value of the vertex.Code:r(h) = (1650 + 190·h)·(480 - 19·h) = 2 - 3610·h + 59850·h + 792000

Calculate the first derivative of r:

If the rebate is 315/38 * 19 = $157.50 that means if the TV set is sold for $322.50 then the revenue has a maximum.Code:r'(h) = 59850 - 7220·h r'(h) = 0 ==> h = 315 ⎯⎯⎯ 38

to c)

the profit is the revenue minus the cost:

Expand all brackets, calculate the first derivative, let the derivative be zero, solve for h and you'll get:Code:p(h) = (1650+190·h)·(480-19·h)-(132000+160·(1650+190·h))

Plug in this value into the function p(h) to get the maximum profit: p_max = $4560625Code:155 h = ⎯⎯⎯ 38

Number of sold TV sets: 2425 pieces.

EB