Dalto Pizza currently sells 1000 pizzas per week at \$18 per pizza. It is planning to reduce the unit price of each pizza. It estimates that for every \$1 discount in price, it can sell 100 more pizzas each week.
(a) Form the weekly revenue function of Dalto Pizza in terms of p, the new unit price of the pizza.
(b) What should the new unit price be in order to maximize weekly revenue? What is the maximum weekly revenue?
My solution is:
$$
R(p)=(18-p)(1000+100p)=-200p^2+800p+18000
$$
$$
R'(p)=-400p+800
$$
let R'(p)=0 and p=2
$$
R''(p)=-400
$$
(which is <0) So p=2 is a maximum point,
Unit Price = 18-2=$16
Sales Volume = 1000+200=$1200
R(max)= $16*$1200=$19200
The solution for weekly revenue function is = 2800-100p^2
p=14
R(max)=$19600
Best Answer
If we let $N(x)$ be the number of units sold at a price of $x$, then the revenue function $R(x)=xN(x)$.
As $N(x)$ is linear, $N(x)=ax+b$.
We have also been told that $N(x-1)=N(x)+100$, or equivalently $N(x-1)-N(x)=100$, and so $a(x-1)-ax=100$, and therefore $a=-100$.
We also know $N(18)=1000=18a+b$ and so we can deduce that $b=2800$.
This gives a function $R()x)=x(-100x+2800)=-100x^2+2800x$.
This differentiates to $-200x+2800$, and has a zero at $x=14$, which is a maximum because the coefficient of $x^2$ is negative.
Therefore the maximum revenue possible is at $x=14$, and $R(14)=14\times N(14)=14\times 1400=19600$.
The method you give uses the substitution $x=18-p$. The price is then calculated as $18-p$, and the number of unit sold is $1000+100p$, which gives a revenue function of $(18-p)(1000+100p)$.
When you multiplied this out, you have a $-200p^2$ term which should be $-100p^2$. You can check algebra online at sites like MathPapa.
Substituting $p=18-x$ into $-100p^2+800p+18000$ gives:
$-32400+3600x-100x^2+14400-800x+18000=2800x-100x^2$.