Find maximum value of the product xy(72−3x−4y) for positive value of x & y.


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Question Number 141164 by iloveisrael last updated on 16/May/21

$F i n d m a x i m u m v a l u e$ $o f t h e p r o d u c t x y (72 - 3 x - 4 y)$ $f o r p o s i t i v e v a l u e o f x & y .$
Commented bymitica last updated on 16/May/21

$m a x = 1152 f o r x = 8, y = 6$
	Answered by EDWIN88 last updated on 16/May/21

	$The terms of the product x, y & 72 - 3 x - 4 y$ $do not have a constant sum . However if we$ $adjust the product and write it as$ $\frac{1}{12} (3 x) (4 y) (72 - 3 x - 4 y)$ $then ignoring the multiplier \frac{1}{12} for a moment$ $we see that the factors 3 x, 4 y & 72 - 3 x - 4 y$ $have a constant 72 . Thus we seek value of$ $x & y satisfying$ $3 x = 4 y = 72 - 3 x - 4 y$ $given the unique values {\begin{cases} x = 8 \\ y = 6 \end{cases}$ $Thus the \max value of xy (72 - 3 x - 4 y) is$ $8 \times 6 \times 24 = 1, 152 ⋇$
	Commented byiloveisrael last updated on 16/May/21

	$Y e s . . . .$
	Answered by mitica last updated on 16/May/21
	$if 72−3x−4y≤0⇒xy(72−3x−4y)≤0 if 72−3x−4y>0⇒(1/(12))∙3x∙4y∙(72−3x−4y)≤^(am−gm) (1/(12))(3x+4y−72−3x−4y)^3 ∙(1/(27))=1152 = for 3x=4y=72−3x−4y⇒x=8,y=6 ⇒max{xy(72−3x−4y)}=1152$
	$i f 72 - 3 x - 4 y ⩽ 0 \Rightarrow x y (72 - 3 x - 4 y) ⩽ 0$ $i f 72 - 3 x - 4 y > 0 \Rightarrow \frac{1}{12} \cdot 3 x \cdot 4 y \cdot (72 - 3 x - 4 y) \overset{a m - g m}{⩽} \frac{1}{12} {(3 x + 4 y - 72 - 3 x - 4 y)}^{3} \cdot \frac{1}{27} = 1152$ $= f o r 3 x = 4 y = 72 - 3 x - 4 y \Rightarrow x = 8, y = 6$ $\Rightarrow m a x {x y (72 - 3 x - 4 y)} = 1152$
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