Struggling while reviewing my old math books. The problem has a prize-function and wants to know how the prize-per-unit should be chosen to maximize the profit at $\mathbf{x=160}$.

First I look analyze the given equation. The given prize function is:

$$ \operatorname{Prize}(x) = 0.003x^3-0.54x^2+96.6x+8100 $$

I calculated the prize-per-unit by dividing by $x$:

$$ \operatorname{PrizePerUnit}(x) = 0.003x^2-0.54x+96.6 + 8100/x $$

First derivative is:

$$ \operatorname{PrizePerUnit}'(x) = 0.006x-0.54 + 8100/x^2 $$

Only real solution is $x=150$ which gives a prize-per-unit of $137.10$.

The usual way to calculate the profit is:

$$ \operatorname{Profit} = \operatorname{Returns} - \operatorname{Prize} $$

If I use the prize-per-unit from the given function:

$$ \operatorname{Profit} = \operatorname{Returns} - \operatorname{Prize} $$

$$ \operatorname{Profit} = x - (0.006x-0.54 + 8100/x^2) $$

I get:

$$ \operatorname{Profit}(x)=-96.6-0.003 x^2+1.54 x-8100/x $$

Which has a maximum at $\mathbf{274.573}$.

And this is where I'm stuck. The solution according to the book is $\mathbf{154.20}$ prize-per-unit. Anybody know how to reach that number?

Integration is apparently not part of the solution, because that is introduced much later in the book. Can this be done with some kind of equivalence-equation?

  • 1
    $\begingroup$ Is this the complete description of the problem? It seems like something is missing. For example, what exactly is $x$? The number of units sold? Unit price? Also, I'm not sure if your profit equation is right -- it would be sale price minus the cost of the product, and I'm not sure what you mean by "returns". Perhaps you could include the exact problem statement in your book? $\endgroup$ – Null Sep 4 '14 at 3:29
  • $\begingroup$ Just to review: in general, to maximize a function $p(x)$ you find the roots of its derivative $p'(x)$, determine if the second derivative $p''(x)$ is negative at those roots (indicating a local maximum instead of minimum), and then find the global maximum by finding which of the local maxima results in the greatest $p(x)$. Once you figure out the correct $p(x)$ you should be able to solve the problem easily. $\endgroup$ – Null Sep 4 '14 at 3:32

You price function gives the total costs to make a number of $x$ products. Call this $C(x)$. Your income is given by the number of sold items times the sale price. If we denote the sale price by $\alpha$, then the income is given by $\alpha x$.

Your profit $P$ can be calculated by subtracting the costs from the income: $$P(x) = \alpha x - C(x).$$ If $P$ has a maximum then it's derivative must be zero there, i.e.: $$P^\prime(x) = \alpha - 0.009x^2 + 1.08x - 96.6 = 0.$$ Now you can simply move all the $x$'s to the right-hand side and substitute $x=160$. Then you will find that $\alpha = 154.20$.

Formally you should check that this indeed gives you a maximum and not a minimum, since we only used that the derivative is zero, which can imply both.


The price function expresses price as a function of quantity offered or demanded. In this specific case, the word "price" refers to the cost of production for $x$ items.

On the other hand, the future income from the sell is described by the "revenue function", which in this case is simply given by the product $jx$, where $j$ is the price at which each item will be sold and $x$ is again the number of items.

The maximal profit is typically obtained by equalizing the first derivatives of the price and revenue function, and checking that the value obtained is actually a maximum and not a minimum. So, to calculate the value of $j$ that maximizes profit for given values of $x$, we have to write

$$\displaystyle j = 0.009 x^2 - 1.08 x+96.6$$

Because the problem asks to find the value of $j$ at $x=160$, we can write

$$\displaystyle j = 0.009 \cdot 160^2 - 1.08 \cdot 160 + 96.6$$

giving $\displaystyle j=154.20$, which can be easily shown to correspond to a maximum.


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