Question

A manufacturer has been selling 1000 flat-screen TVs a week at $500 each. A market survey indicates that for A manufacturer has been will increase by 100 per week. each $10 rebate offered to the buyer, the number of TVs sold
a) Find the demand function (price rho as a function of units sold x).

(b) How large a rebate should the company offer the buyer in order to maximize its revenue?

(c) If its weekly cost function is C(x) 68,000+100x, how should the manufacturer set the how should the manufacturer set the size of the rebate in order to maximize its profit?

Answer 1

Answer:

a) Demand function: $q=6000-10\cdot p$

b) The rebate should be of $200, so the sale price becomes $300 per unit.

c) The rebate should be of $150, so the sale price becomes $350 per unit.

Step-by-step explanation:

a) In this case, we have a known point of the demand function (1000 units sold at $500), and the slope of a linear function (increase by 100 units fora decrease in $10).

We can express the demand function (linear) as:

$q=b+m\cdot p$

To calculate the slope m, we use:

$m=\Delta q/\Delta p=(+100)/(-10)=-10$

To calculate b, we use the known point and the calculated slope:

$q=b+m\cdot p\\\\b=q-m\cdot p=1000-(-10)\cdot (500)=1000+5000=6000$

Then the demand function is:

$q=6000-10\cdot p$

b) The revenue can be expressed as:

$R=q\cdot p = (6000-10p)\cdot p=6000p-10p^2$

To maximize, we can derive and equal to zero

$dR/dp=6000-2*10p=0\\\\20p=6000\\\\p=300$

The rebate should be of $200, so the sale price becomes $300 per unit.

c) If we take into account the cost, we have that

$R=q\cdot p-C=(6000p-10p^2)-(68000+100q)\\\\R=(6000p-10p^2)-(68000+100(6000-10p))\\\\R=6000p-10p^2-(68000+600000-1000p)\\\\R=-10p^2+7000p-668000$

To maximize, we can derive and equal to zero

$dR/dp=-20p+7000=0\\\\p=7000/20=350$

The rebate should be of $150, so the sale price becomes $350 per unit.