The sum of two numbers is less than $120$ with he first number is bigger than the second. The differences (subtraction) between the two numbers is $60$. Find the lowest result of multiplication between the two numbers.

This is an easy question, but I can't seemed to solve this one. Anyway, from the question we obtained:

$$ a+b<120$$ $$a-b=60$$ And $$a>b$$

As if it normal equations, I would have solve them easily. Any suggestions?

  • $\begingroup$ $a,b$ are integers or naturals or else? $\endgroup$ – Error 404 Mar 4 '17 at 13:43
  • $\begingroup$ @VikrantDesai they are integers. $\endgroup$ – Adola Mar 4 '17 at 13:50

We want to minimize $ab$. Since $a-b = 60 \implies a=60+b$, then minimizing $ab$ is the same as minimizing $(60+b)b$.

Using calculus, we can find any local minimums by setting the derivative of this equal to $0$ and solving for $b$, $$\frac{d}{dx} (60+b)b = 0$$ $$\frac{d}{dx} 60b+b^2 = 0$$ $$ 60+2b = 0$$ $$b = -30$$

And thus $a = 60+b = 60-30 = 30$.

And thus the multiplication of the two is minimized at $ab = (30)(-30) = -90$.

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