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Needless to say that what I've already tried it didn't work hence I'm requesting assistance on this. Overall I want to know if there is some graphical approach on this problem. Can this be expressed as the intersection sector between two lines?. I'm not sure if this applies here so guidance on that part would be greatly appreciated.

The problem is as follows:

At a bakery shop in Taipei a chef can prepare different kinds of high end gourmet pies. His new creation is a blueberry cake which costs $\textrm{60 USD}$ to make. If he sells them at $x\,\textrm{USD}$ per cake $\left(60\leq x \leq 130\right)$. He estimates the quantity he can sell to be $\left(130-x\right)$ a month. If the maximum profit depends solely on the price he sells that cake. Find the sale price of each cake so that his monthly profit be the maximum.

\begin{array}{ll} 1.&85\,\textrm{USD}\\ 2.&100\,\textrm{USD}\\ 3.&95\,\textrm{USD}\\ 4.&80\,\textrm{USD}\\ 5.&90\,\textrm{USD}\\ \end{array}

What I tried was to use the inequation given in the problem. As I felt that the intended answer would come from it. Therefore I used it as follows:

$$60\leq x \leq 130$$

Multiplying by $-1$ so I could obtain the $-x$:

$$-60 \geq -x \geq -130$$

Summing $130$:

$$130-60 \geq 130-x \geq 0$$

$$70 \geq 130 - x \geq 0$$

Since it mentioned that for his profit to the maximum this ammount must be the quantity he can sell that gourmet cake a month. Then I figured it to be $70\,\textrm{USD}$. But this doesn't appear in the alternatives.

Therefore I'm left dumbfounded on what could I be missing?. There is also the part of the inequation which I am not sure to be translating it correctly. Is that the correct way?. I feel that I need to subtract the cost to make each cake so what is left is the profit, to which I felt did with the inequation.

Overall can somebody help me with this?. I'm stuck at this part. To me an answer which would help me the most is one which could include some explanation or perhaps telling which part I did not understood correctly.

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  • $\begingroup$ What is your reasoning for thinking that he should sell the cakes at a price which is the maximum number of cakes he can sell in a month? $\endgroup$
    – user445909
    Commented Feb 13, 2019 at 20:36
  • $\begingroup$ @E-mu My reasoning for that was by following with what was given in the interval. Hence relate those quantities. I may not understood that part correctly. Again, I didn't knew how to use that information to solve the problem. But the more I think on it, it doesn't make sense as both quantities cannot be the same. $\endgroup$ Commented Feb 13, 2019 at 21:17

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I can't think of a reason why the price he should sell each cake at is the maximum number of cakes he can sell in a month. You could solve this problem by finding an expression for the profit he makes in a month in terms of the selling price $x$, and then maximise the profit with respect to $x$.

We assume:

  • He sells $130 -x$ blueberry cakes in a month.
  • The ingredients for each cake cost $\$60$.
  • He sells each cake for $\$x$.
  • He does not waste ingredients, in the sense that all of the ingredients he buys are used up in making cakes. This means that the number of cakes he sells is the number of cakes he buys ingredients for.

The profit $P$ he would make from the cakes in a month is then given by \begin{align}P &= (\text{selling price of each cake})\cdot(\text{# cakes sold}) - (\text{cost of ingredients for each cake})\cdot (\text{# of cakes he buys ingredients for})\\ &= x(130-x) - 60(130-x). \end{align} The resulting profit is quadratic in $x$. Can you find $x$ such that $P$ is maximised?

The only graphical approach I can think of, is one to find the solution from here (spoiler):

In the quadratic expression, the coefficient of $x^2$ is negative, so you know that the corresponding parabola in the $x$-$P$ plane will have a maximum at the vertex. The $x$-coordinate of the vertex is in the middle of the two roots, which a quick factorisation of $P$ tells us are $60$ and $130$. The selling price which maximises profit is therefore given by $x = \frac{60 + 130}{2} = 95$.

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    $\begingroup$ This problem was found in a precalculus book therefore should been able to solve without derivatives. Although just at the time you modified your answer I did also come up with the spoiler you are hiding. $\endgroup$ Commented Feb 13, 2019 at 21:18
  • $\begingroup$ It is not shown in the answer but if I go on the derivatives route this would result into $P(x)=-x^2+190x-7800$ so this becomes into $P'(x)=-2x+190$ and equating the last to zero (to find the maximum) would result into $x=95$ which corresponds to what you had wrote as the answer in the spoiler in other words the sale price. But the maximum profit is not that amount but rather the result found evaluated in the function, which is $P(95)=95\left(130-95\right)-60\left(130-95\right)=1225$ and those $1225\$$ would be the maximum profit right?. $\endgroup$ Commented Feb 13, 2019 at 21:30
  • $\begingroup$ I did also drawn a sketch on my notebook of the parabola which you mentioned given the points $60$ and $130$ but I didn't realized that half that point was the maximum until I used the derivatives, so the approach of taking the average between those seems more faster than to taking the derivative of the function. Overall I overlooked the equation for the profit. Maybe it is obvious but I didn't realized that the cakes he sell is the number of cakes he makes and that's how this problem is solved. $\endgroup$ Commented Feb 13, 2019 at 21:35
  • $\begingroup$ The inequation given at the beginning was misleading as it made me to believe that this should be used to find the maximum profit. Does it exist a way to avoid this source of confusion for similar problems?. $\endgroup$ Commented Feb 13, 2019 at 21:37
  • $\begingroup$ Your calculation of the maximum profit is correct. However the problem, as stated, asks for the selling price at which the maximum profit is attained, which is $\$95$. $\endgroup$
    – user445909
    Commented Feb 13, 2019 at 22:08

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