Find the maximum value of $x^2y$ given constraints Find the maximum value of $${ x }^{ 2 }y$$ subject to the constraint $$x+y+\sqrt { 2{ x }^{ 2 }+2xy+3{ y }^{ 2 } } =k$$ where k is a constant.
I tried it by substituting value of x and then differentiating w.r.t $x$ but not able to proceed further.
 A: Let $y=a x$ and from the constraint (assuming $x>0$
$$x=\frac{k}{\sqrt{3 a^2+2 a+2}+a+1}\qquad y=\frac{ak}{\sqrt{3 a^2+2 a+2}+a+1}$$
$$x^2y=\frac{a k^3}{\left(\sqrt{3 a^2+2 a+2}+a+1\right)^3}$$ Differentiate with respect to $a$ to get
$$-\frac{(2 a-1) \left(3 a+2+\sqrt{a (3 a+2)+2}\right) k^3}{\sqrt{a (3 a+2)+2}
   \left(a+\sqrt{a (3 a+2)+2}+1\right)^4}=0$$
So, either $a=\frac 12$ which would give
$$x^2 y=\frac{4 k^3}{\left(3+\sqrt{15}\right)^3}$$ or
$a=-\frac{1}{6} \left(5+\sqrt{13}\right)$ which would give
$$x^2y=\frac{1}{54} \left(35-13 \sqrt{13}\right) k^3$$
A: Hint:
For convenience, turn the original constraint to a quadratic form.
$$2x^2+2xy+3y^2-(k-x-y)^2=0$$
or
$$x^2+2kx+2y^2+2ky-k^2=0.$$
Thus the equations deduced with Lagrangian multipliers are
$$\begin{cases}2xy=\lambda(2x+2k),\\x^2=\lambda(4y+2k).\end{cases}$$
By elimination,
$$2x^2+2kx-8y^2-4ky=0$$ and you need to intersect two conics.
A: $$g(x,y)=x+y+\sqrt { 2{ x }^{ 2 }+2xy+3{ y }^{ 2 } }= x+y + Q(x,y) = k \tag1$$
for the object function
$$f(x,y)=x^2y $$
By Lagrange multiplier method
$$\dfrac{g_x}{g_y}= \dfrac{f_x}{f_y} $$
$$\dfrac{2y}{x}=1+\dfrac{(4x+2y)}{2Q}=1+\dfrac{(2x+6y)}{2Q}$$
simplifying
$$Q(2y-x)= 2x^2-xy-6y^2$$
has a common factor $ (2y-x)$ to cancel
$$=(x-2y)(2x+3y)\rightarrow Q=(2x+3y)$$
Squaring
$$ Q^2=4x^2+12 x y+9y^2=2x^2+10 xy+6y^2$$
simplifying to find roots of quadratic
$$x^2+5 x y+3 y^2=0;\quad  \dfrac{y}{x}=\dfrac{-5\pm \sqrt{13}}{6}; $$
which are a pair of straight lines.
If $(p,q)$ are these roots say $ y=px,\;y=qx\;$  Plug in the first of two roots into (1)
$$ k = x( 1+p+ \sqrt{2+2p+3p^2}) =C x\;$$ say, then the maximum value is
$$x^2y=x^3\cdot\dfrac{y}{x}=p  k^3/C^3=\dfrac{pk^3}{( 1+p+ \sqrt{2+2p+3p^2})^3}$$
where
$$p=\dfrac{\sqrt{13}-5}{6}$$
The minimum value can be found by $y=qx$ in a similar fashion.
A: One could proceed by the method of Lagrange multipliers, you can find more information on this either at:

*

*https://en.wikipedia.org/wiki/Lagrange_multiplier,

*or at the following links which houses a number of examples https://math.berkeley.edu/~scanlon/m16bs04/ln/16b2lec3.pdf.

Now, we aim to maximise the function $f(x,y) = x^2 y$ subject to the constraint $g(x,y) = 0 $ where
$$g(x,y) = x + y + \sqrt{2x^2 + 2xy + 3y^2} - k $$
for some constant $k \in \mathbb{R}$. The constraint in the above form can be simplified (as to remove the square roots) by simply squaring it. So now instead consider the function
$$h(x,y) = x^2 + 2y^2 + 2k(x+y) - k^2.$$
Then the above problem is equivalent to maximising $f(x,y)$ with respect to the constraint $h(x,y) = 0$.
Setting
$$L(x,y,\lambda) := f(x,y) + \lambda g(x,y),$$
and solving the system of equations
$$\frac{\partial L}{\partial x}(x,y,\lambda) = 0, \quad \frac{\partial L}{\partial y}(x,y,\lambda) = 0, \quad \frac{\partial L}{\partial \lambda}(x,y,\lambda) = 0$$
will yield the required solution.
Can you take it from here?
