if-x-2-y-2-16-2-y-2-z-2-24-2-z-2-t-2-42-2-and-t-2-x-2-38-2-find-max-x-z-y-t-

Question Number 151315 by mathdanisur last updated on 20/Aug/21

if x^{2} + y^{2} = 16^{2}; y^{2} + z^{2} = 24^{2}

z^{2} + t^{2} = 42^{2} and t^{2} + x^{2} = 38^{2}

find \max [(x + z) (y + t)] = ?

Answered by dumitrel last updated on 20/Aug/21

(x + z) (y + t) = x y + z t + x t + z y

{(x y + z t)}^{2} ⩽ (x^{2} + t^{2}) (y^{2} + z^{2}) = 38^{2} \cdot 24^{2}

\Rightarrow x y + z t ⩽ 38 \cdot 24

x y + z t = 38 \cdot 24 \Leftrightarrow \frac{x}{y} = \frac{t}{z}

{(x t + z y)}^{2} ⩽ (x^{2} + y^{2}) (t^{2} + z^{2}) = 16^{2} \cdot 12^{2}

\Rightarrow x t + y z ⩽ 16 \cdot 12

x t + y z = 16 \cdot 12 \Leftrightarrow \frac{x}{t} = \frac{y}{z} \Rightarrow

(x + z) (y + t) ⩽ 1104

m a x (x + z) (y + t) = 1104 f o r

x^{2} = \frac{5120}{33}; y^{2} = \frac{3328}{33}; z^{2} = \frac{15680}{33}; t^{2} = \frac{42532}{33}

Commented by mathdanisur last updated on 20/Aug/21

Thank You S er, cool