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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
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  • $\begingroup$ Set $N(x)$ as the Number of pizzas sold, where $x$ is the price. We know that $N(18)=1000$. With this notation, how would you write "for every \$1 discount in price, it can sell 100 more pizzas"? $\endgroup$
    – Matti P.
    May 14, 2020 at 8:15
  • $\begingroup$ Check my working up $\endgroup$
    – Big_Smoke
    May 14, 2020 at 8:21
  • $\begingroup$ How did you come up with $R(p) = (18-p)(100+100p)$ ? I would approach this like such: Assume linear function for $N(x)$: $N(x) = ax+b$. We know that $N(18) = 1000$ and $N'(x) = -100 = a$. Therefore, we can solve for $a$ and $b$: $N(x) = -100x + 2800$. To get the revenue, we multiply this by $x$: $R(x) = xN(x)$ ... $\endgroup$
    – Matti P.
    May 14, 2020 at 9:28
  • $\begingroup$ ibb.co/8jFZPtW $\endgroup$
    – Big_Smoke
    May 14, 2020 at 9:35
  • $\begingroup$ Check the above link $\endgroup$
    – Big_Smoke
    May 14, 2020 at 9:35

1 Answer 1

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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:

$-32400+3600x-100x^2+14400-800x+18000=2800x-100x^2$.

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  • $\begingroup$ Thanks buddy. Appreciate your help. I'm just dumb tbh my brain is filled with fog. I could not think due to potential loss of brain cells and PTSD. $\endgroup$
    – Big_Smoke
    May 14, 2020 at 19:00

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