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Problem to solve:

To buy 2 products the seller gives 30% discount for the less expensive product. The customer/buyer payed for both products 300$. What is the biggest/max price of the less expensive product before the 30% discount?

What I did:

let x be the first product

let y be the second product which is lowered by 30%, the less expensive product

First equation: $x+y-(y*0.3)=300{$}$

Second equation ? I tried but just did a modification of the previous one which is incorrect. Also would derivatives apply here? $$$$

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You have shown that $$x+\frac{7}{10}y=300$$ and so $$x=300-\frac{7}{10}y$$ and, in order for $y$ to be less expensive, $$y \lt x$$ by substitution, $$y \lt 300-\frac{7}{10}y$$ $$\frac{17}{10}y \lt 300$$ $$y \lt \frac{3000}{17} \approx 176.48$$ and so there is no real maximum value of $y$, since it cannot equal or exceed $\approx \$176.48$ but can be any value arbitrarily close to it and less than it.

The inequality $$y \lt x$$ was really the "second equation" that you were looking for, but it isn't an equation.

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  • $\begingroup$ damn, so easy when you showed it. Tnx. Would overload my mind too much If I hadn't asked the question. $\endgroup$ – eugene_sunic Jul 3 '17 at 16:12
  • $\begingroup$ @eugensunic No problem, don't worry about it. If it helped, just $\checkmark$! :D $\endgroup$ – Franklin Pezzuti Dyer Jul 3 '17 at 16:15
  • $\begingroup$ I'm waiting for the interval to pass to +15, already +10 ! $\endgroup$ – eugene_sunic Jul 3 '17 at 16:16
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This is an interesting question.

Here's a more direct approach.

The maximum pre-discount price $y^*$ of $Y$ (the less expensive product) must be equal to $x$, the price of $X$ (the more expensive product).

If ${y^*}'$ is the maximum post-discount price of $Y$, then ${y^*}':x=7:10$ which gives $$y^*=x=\frac {10}{17}\cdot 300=\color{red}{176.47}$$

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  • $\begingroup$ Nice approach, but $\frac{3000}{17} = 176.47$ to the nearest hundredth. $\endgroup$ – N. F. Taussig Jul 4 '17 at 8:53
  • $\begingroup$ @N.F.Taussig - Thanks for pointing out the typo! :) $\endgroup$ – hypergeometric Jul 4 '17 at 10:49

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