if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$
putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$
i wan,t be able to go further,could some help me with this
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Is it $x^2y+z=4$? – Shrey Aryan Nov 22 '16 at 06:03
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Could we do this using calculus? Using Lagrange multipliers? With that, we find that it's sufficient to find $x,y,z,\lambda$ such that $$ y + z = \lambda\ x + z = 2\lambda\ y + z = \lambda\ x + 2y + z = 4 $$ – Ben Grossmann Nov 22 '16 at 06:11
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Since $z=4-x-2y$, $$\begin{align}xy+yz+zx&=xy+y(4-x-2y)+(4-x-2y)x\\&=xy+4y-xy-2y^2+4x-x^2-2xy\\&=-x^2+(4-2y)x-2y^2+4y\\&=-(x^2+(2y-4)x)-2y^2+4y\\&=-\left((x+y-2)^2-(y-2)^2\right)-2y^2+4y\\&=-(x+y-2)^2+(y-2)^2-2y^2+4y\\&=-(x+y-2)^2-y^2+4\\&\le 4\end{align}$$ The equality is attained when $x+y-2=y=0$, i.e. $x=2,y=0,z=2$.
mathlove
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$$xy+yz+zx+y^2 = (x+y)(y+z) \le \frac{(x+2y+z)^2}{4} = 4$$
$$\Rightarrow xy+yz+zx \le 4-y^2 \le 4$$
Equality occurs when $y=0$, $x=z=2$
juantheron
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Hari Shankar
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$$xy+z(x+y)=xy+(4-x-2y)(x+y)=4x+4y-x^2-2y^2-2xy=k\text{(say)}$$
$$x^2+2x(y-2)+2y^2-4y-k=0$$
As $x$ is real, the discriminant $$4(y-2)^2-4(2y^2-4y-k)\ge0$$
The equality occurs if $x=-\dfrac{2(y-2)}2$
$$\iff k\le4-y^2\le4$$ The equality occurs if $y^2=0$
lab bhattacharjee
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