Let a, b, c be the three sides of a triangle $$△ABC$$, If the area $$S$$ and the sides satisfy[[]] $$ S = a^2-(b -c)^2 $$, and $$b+c = 8$$, find the maximum value of $$S$$

Question

Let a, b, c be the three sides of a triangle $$△ABC$$, If the area $$S$$ and the sides  satisfy[[]] $$ S = a^2-(b -c)^2 $$, and  $$b+c = 8$$, find the maximum value of  $$S$$

Uniteasy Admin · Accepted Answer

Using the formula for the area of a triangle
$$S = a^2-(b -c)^2 = \dfrac{1}{2}bc\sin A $$
According to Law of cosines
$$ a^2-(b -c)^2 = 2bc - 2bc\cos A$$
$$∴ 1 - \cos A = \dfrac{1}{4}\sin A $$
$$\dfrac{1 - \cos A}{\sin A} = \dfrac{1}{4}$$
$$	an \dfrac{A}{2} = \dfrac{1}{4}$$
$$∴ \sin A = \dfrac{2	an \dfrac{A}{2} }{1+	an^2 \dfrac{A}{2} } $$
$$= \dfrac{\dfrac{1}{2} }{1+\dfrac{1}{16} } = \dfrac{8}{17} $$
$$S =\dfrac{1}{2}bc\sin A = \dfrac{4}{17}bc = \dfrac{4}{17}b(8-b)$$
$$= -\dfrac{4}{17}[(b-4)^2-16 ]$$
Therefore,
when $$b = 4$$, $$S_{max} = \dfrac{64}{17} $$

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