Find the maximum value of $x^2y$ given constraints

Question

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.

Answer 1

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?

Answer 2

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 $$x^2y=\frac{1}{54} \left(35-13 \sqrt{13}\right) k^3$$

Answer 3

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.

Answer 4

$$g(x,y)=x+y+\sqrt { 2{ x }^{ 2 }+2xy+3{ y }^{ 2 } }= x+y + Q(x,y) = k \tag1$$

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.