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A school class is saving money for a classtrip and therefore sell cakes. The function $f(x)=x(x-25)(x-15)$ describes how much money the class saves in total for selling cakes.

f(x) is the total amount of sum in dollars and x is what they earn per cake. Their teachers said that they cant sell a cake for more than 15 dollars but not more. How much money should they ask per/cake to earn as much as possible?

This is what I did: $$x(x^2-15x-25x+375)$$ $$x(x^2-40x+375)$$ $$x^3-40x^2+375x$$ Then derivate $= 3x^2-80x+375$. Then $3x^2-80x+375=0$ to find maximum but I cant divide $80/3$, what more can I do to answer the question?

TYZ
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Ilovemath
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1 Answers1

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So we have a function $f(x) = x(x-15)(x-25)$. We need to maximize this function under the constraint that $x \le 15$. Obviously for $x\le0$ $f(x)$ is negative so our constraints are, $0\le x\le15$.

Now, consider $f(x) = x^3 -40x^2 + 375x \implies f'(x) = 3x^2 - 80x + 375$

We want to find a root of $f'(x)$, i.e, solve for $f'(x)=0$, such that $0\le x\le 15$.

Applying the quadratic formula,

$$x = \frac{80 \pm \sqrt{6400 - 4500}}{6}$$

Simplifying this becomes, $$x = \frac{40 - 5\sqrt{19}}{3}$$ and $$f(x) = \frac{250}{27}\left(28 + 19\sqrt{19}\right)$$

This is the maximum value. I omitted the calculation, I think you can do that yourself?

Guy
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  • Yes, do I only have to calculate 250/27 and can I skip 28 + 19... or do I have to multiply them both? – Ilovemath Mar 12 '14 at 11:04
  • You have to calculate $250/27$ and $28 + 19\sqrt{19}$. And multiply the two results. In case approximate values will do, $f(x)=1026$ at $x=6$. The absolute maximum for $f(x)$ is around $1026.1$. So I think the approximation is good enough? – Guy Mar 12 '14 at 11:08
  • Yes, thanks for you effort and time! Very kind of you. – Ilovemath Mar 12 '14 at 11:10
  • I am very happy to help. You can 'accept' my answer if it is good enough. – Guy Mar 12 '14 at 11:10
  • Ok, I will when I register – Ilovemath Mar 12 '14 at 11:11
  • Oh, SE now allows non registered users to have assign names? I didn't know that. Anyway bye now. Feel free to post a question again, in case you need help. :) – Guy Mar 12 '14 at 11:12