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Please help solving this problem :)

A restaurant has a fixed price of $36 for a complete dinner. The average number of customers per night is 200. The owner estimates that for each dollar increase in the price of the dinner, there will be, on average, four fewer customers per night. Estimate the price for the dinner that will produce the maximum amount the owner can expect to receive

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  • $\begingroup$ Do you know how to maximize a quadratic function? $\endgroup$ – Andrew Chin Sep 21 '19 at 16:35
  • $\begingroup$ Assume the increase in price as x and find the total money received by the owner in terms of x first. Then, you just need to maximize it $\endgroup$ – Rishabh Jain Sep 21 '19 at 16:41
  • $\begingroup$ Similar to: math.stackexchange.com/questions/2294416/the-greatest-revenue/… $\endgroup$ – NoChance Sep 21 '19 at 18:49
  • $\begingroup$ The key here is to derive the revenue function. The first part of this video gives an acceptable explanation of how to handle such a question. $\endgroup$ – NoChance Sep 21 '19 at 20:44
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Let $x$ represent the unit of change to price and customer expectancy.

Define the revenue function based on the relationship $\textrm{revenue = price} \times \textrm{volume}$. We have

$$R(x)=(36+x)(200-4x)$$

Since this is a quadratic function, we can find the maximum by converting the equation into vertex form.

\begin{align} R(x)&=(36+x)(200-4x)\\ &=7200+56x-4x^2\\ &=-4(x^2-14x)+7200\\ &=-4(x^2-14x\color{blue}{+49-49})+7200\\ &=-4(x^2-14x+49)+196+7200\\ &=-4(x-7)^2+7396 \end{align} Since the vertex to $R(x)$ is at $(7,7396)$, that means the maximum when $x=7$. Substitute this into the price expression $$36+x=36+(7)=43.$$ Therefore, the price that would produce the maximum revenue is $\$43$.

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  • $\begingroup$ This method does give you more information than is required, so a more concise method could just be to find the equation of the axis of symmetry of the function (in this case, it would be $x=7$, and there are many, many ways you can come about this answer) and use it to solve the problem. $\endgroup$ – Andrew Chin Sep 21 '19 at 16:47
  • $\begingroup$ Could you please clarify how did you come with the first quadratic equation? $\endgroup$ – NoChance Sep 21 '19 at 18:43
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    $\begingroup$ For a single price, revenue can be found by taking the product of price and volume. For no change in price or volume, the revenue is simply $36\times200$. But, since price and volume changes proportionally, we can use a single variable, $x$, to represent a proportion. Numerically speaking, every time the price increases by $1$ dollar, the number of customers drop by $4$. So if we increase the price by $x$, the number of customers decrease by $4x$. $\endgroup$ – Andrew Chin Sep 21 '19 at 18:58
  • $\begingroup$ Thank you for the explanation. I still can't see the quadratic formula. I will study it further. $\endgroup$ – NoChance Sep 21 '19 at 20:25
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I think I messed up somewhere, but I'll give it a try:

The total money, given that the price increased by $n$ dollars, should be $$(36+n)(200-4n)=-4n^2+56n+7200.$$

Now, we have to maximize this. This makes sense to maximize because $a$ is negative. Now, for a general quadratic equation $ax^2+bx+c$ where $a<0$, the maximum occurs at $$x=-\frac b{2a}\text{ is: }c-\frac{b^2}{4a}$$,

$$ and you should be able to get your answer from here.

Hope this helps.

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    $\begingroup$ For the proof of the maximization, we simply use completing the square. We have $$\begin{align*}ax^2+bx+c=0 \\ x^2+\frac ba x + \frac ca = 0 \\ \left(x+\frac b{2a}\right)^2+\frac ca-\frac{b^2}{4a^2}=0\\ a\left(x+\frac b{2a}\right)^2+c-\frac{b^2}{4a}=0.\end{align*}$$ To maximize this, we must have $a\left(x+\frac b{2a}\right)^2=0$, so the maximum occurs at $$x=-\frac b{2a}\text{ and is }c-\frac{b^2}{4a}.$$ $\endgroup$ – guest Sep 21 '19 at 17:14
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If the meal costs $36 + k$ dollars the owner will have $200 - 4k$ costumers.

So he will make $(36 + k)(200 - 4k)$ dollars.

What is the highest possible value of $(36+k)(200-4k)$?

.....

$(36+k)(200-4k) = 7200 + 56k - 4k^2=$

$7200 - (4k^2 -56k) = $

$7200 - ((2k)^2 - 2*2k*14) =$

$7200 + 32^2 - ((2k)^2 - 2*2k*32 + 14^2) =$

$7200 + 196 - (2k - 14)^2 =$

$7396 - (2k-14)^2 \le 8224$.

So if $(2k-14)^2 > 0$ we will have that $7396-(2k-14)^2 < 7396$.

If $(2k-32)^2 = 0$ we will have that $7396-(2k-14)^2 = 7396$.

If $(2k-14)^2 < 0$ we will have broken the laws of space and time and will all die horribly, alone, and in pain.

So the maximum profit revenue is $7396$ when $2k -14=0$.

SO if $k =7$ and he charges $36+7 = 43$ he will make the most money.

But he will lose $28$ costumers. Hard luck for them I guess..... All for just $196$.... Oh, $196$ per night is a significant amount of money. To heck with those $28$ cheapskates...

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