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)]=?


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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

	$Thank You S er, cool$
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