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-

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 {\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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