# How to solve this optimisation problem?

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

• 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"? May 14, 2020 at 8:15 • Check my working up May 14, 2020 at 8:21 • 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)\$ ... May 14, 2020 at 9:28
• ibb.co/8jFZPtW May 14, 2020 at 9:35
• Check the above link May 14, 2020 at 9:35

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:

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

• 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. May 14, 2020 at 19:00