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Find the largest possible area of the triangle $ABC$ such that $AB$ ≤ 2, $BC$ ≤ 3, and AC ≤ 4.

Note: The solution should not contain any calculus method

Can someone tell me how to maximize the area of a triangle? Do we have to make the sides as close as possible to each other to get the maximum area?

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Since $AB$ and $BC$ have lower upper bounds, they should be set to maximum. Then if $AC$ is also set to maximum, angle $B$ becomes blunt ($2^2+3^2<4^2$), which is not good. So we set angle $B$ to $90^\circ$ and the maximum area should be $\frac{1}{2}\times 2\times 3=3$. Once the maximum is found (guessed), it's not hard to prove it:

$$S=\frac{1}{2}|AB||BC|\sin B\leq\frac{1}{2}|AB||BC|\leq\frac{1}{2}\times 2\times 3=3.$$

No calculus. Just a simple trigonometry.

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  • $\begingroup$ Great solution! Very clear! Appreciate it! $\endgroup$ – Nariman Zendehrooh Oct 21 '17 at 4:04
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you can also use Heron's formula. Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is :- Put the upper limits of a=2,b=3,c=4

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  • $\begingroup$ Great formula! But, why don't we get the exact value of 3 by using this formula? We would get 2.9. $\endgroup$ – Nariman Zendehrooh Oct 21 '17 at 4:03
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    $\begingroup$ We've got to put in $a=2, b=3, c=\sqrt{a^2+b^2}=\sqrt{13}$ to get the same answer. To find why $c=\sqrt{13}$ maximizes the area requires calculus, if no geometry is relied on. Or maybe completing the square could do it. $\endgroup$ – Zhuoran He Oct 21 '17 at 4:06
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    $\begingroup$ Yes. Completing the square works. $$16S^2=4a^2b^2-(a^2+b^2-c^2)^2\leq 4a^2b^2.$$ Therefore the maximum area is $S=\frac{1}{2}ab=3$. $\endgroup$ – Zhuoran He Oct 21 '17 at 4:17
  • $\begingroup$ yes, Heron's formula will give the approx answer. here is why :- jwilson.coe.uga.edu/EMT725/Heron/Heron.html $\endgroup$ – Koshinder Oct 21 '17 at 4:21

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