Let a,b,c be positive constants. Among all real number x and y satisfying ax+by=c , find the maximum value of product xy.


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Question Number 138797 by bramlexs22 last updated on 18/Apr/21

$L e t a, b, c b e p o s i t i v e c o n s t a n t s .$ $A m o n g a l l r e a l n u m b e r x a n d y$ $s a t i s f y i n g a x + b y = c, f i n d t h e$ $m a x i m u m v a l u e o f p r o d u c t$ $x y .$
	Answered by TheSupreme last updated on 18/Apr/21

	$i f b = 0 o r a = 0 x y i s u n l i m i t e d$ $e l s e : y = - \frac{a}{b} x - \frac{c}{b}$ $\frac{a}{b} = m$ $\frac{c}{b} = q$ $y = - m x - q$ $σ = x y = x (m x + q) = - {m x}^{2} - q x = - x (m x + q)$ $\frac{\partial σ}{\partial x} = - 2 m x - q = 0 \to x = - \frac{q}{2 m}$ $σ_{m a x} = \frac{q}{2 m} (\frac{q}{2}) = \frac{q^{2}}{4 m} = \frac{c^{2}}{4 a b}$
	Answered by EDWIN88 last updated on 18/Apr/21
	$without calculus. let { ((ax = u)),((by = v)) :} ⇒ u+v = c ; so u.v minimum where { ((u=(c/2))),((v=(c/2))) :} then (uv)_(min) = (c^2 /4) ⇒(ax.by)_(min) = (c^2 /4) or xy_(min) = (c^2 /(4ab)).$
	$without calculus .$ $let {\begin{cases} ax = u \\ by = v \end{cases} \Rightarrow u + v = c; so u . v minimum$ $where {\begin{cases} u = \frac{c}{2} \\ v = \frac{c}{2} \end{cases} then {(uv)}_{\min} = \frac{c^{2}}{4}$ $\Rightarrow {(ax . by)}_{\min} = \frac{c^{2}}{4} or {xy}_{\min} = \frac{c^{2}}{4 ab} .$
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