Question

How do you maximize and minimize $f\left( x,y \right)=x-x{{y}^{2}}$ constrained to $0\le {{x}^{2}}+y\le 1$ ?

Answer 1

We need to find the local maxima and local minima for the given question. We start to solve the problem by assuming two slack variables. Then, we use the Lagrange multipliers technique to get the required result.

Complete step by step solution:
We are given a function in x, y and need to find the local maxima, local minima of the same. We will solve the question by using the Lagrange multipliers technique.
A slack variable is used to transform an inequality constraint into an equality. The transformation is done by adding a value on the left-hand side of the inequality. The value of the slack variable cannot be negative. Its value can be either positive or zero.
Let us consider two slack variables a, b.
The given function is $f\left( x,y \right)=x-x{{y}^{2}}$
Now, According to Lagrange multiplier technique,
$\Rightarrow {{g}_{1}}\left( x,y,a \right)={{x}^{2}}+y-{{a}^{2}}-1=0$
$\Rightarrow {{g}_{2}}\left( x,y,b \right)={{x}^{2}}+y-{{b}^{2}}-1=0$
Forming the function from the above, we get,
$\Rightarrow L\left( x,y,a,b,{{\lambda }_{1}},{{\lambda }_{2}} \right)=f\left( x,y \right)+{{\lambda }_{1}}{{g}_{1}}\left( x,y,a \right)+{{\lambda }_{2}}{{g}_{2}}\left( x,y,b \right)$
The local maxima and local minima can be found by solving further,
$\Rightarrow \nabla L\left( x,y,a,b,{{\lambda }_{1}},{{\lambda }_{2}} \right)=\overrightarrow{0}$
The above expression can also be written as,
$\Rightarrow 1+2\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)x-{{y}^{2}}=0$
$\Rightarrow {{\lambda }_{1}}+{{\lambda }_{2}}-2xy=0$
$\Rightarrow {{a}^{2}}-{{x}^{2}}-y=0$
$\Rightarrow {{\lambda }_{1}}a=0$
$\Rightarrow {{b}^{2}}+{{x}^{2}}+y-1=0$
$\Rightarrow {{\lambda }_{2}}b=0$
Now, we have to solve for the values of $x,y,a,b,{{\lambda }_{1}},{{\lambda }_{2}}$
The five values obtained for each of the variables are given as follows,
$x=0,-1.09545,1.09545,-0.66874,0.66874;$
$y=1,-0.2,-0.2,-0.447214,-0.447214;$
$a=-1,-1,-1,0,0;$
$b=0,0,0,-1,-1;$
${{\lambda }_{1}}=0,0,0,0.59814,-0.59814;$
${{\lambda }_{2}}=0,0.438178,-0.438178,0,0$
The above points are required to be qualified.
The first and third values in the above activate constraint ${{g}_{2}}\left( x,y,0 \right)$
The second and fourth values in the above activate constraint ${{g}_{1}}\left( x,y,0 \right)$
From the above,
$\Rightarrow f\circ {{g}_{1}}\left( x \right)=x-{{x}^{5}}$
$\Rightarrow f\circ {{g}_{2}}\left( x \right)=-{{x}^{3}}\left( {{x}^{2}}-2 \right)$
Substituting the values of $x$ in the above equation,
$\Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f\circ {{g}_{1}}\left( -0.66874 \right) \right)=5.9814$
The point is a local minimum
$\Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f\circ {{g}_{1}}\left( 0.66874 \right) \right)=5.90567$
The point is a local maximum
$\Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f\circ {{g}_{2}}\left( 0 \right) \right)=0$
The point is non-decidable whether it is local minima or maxima
$\Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f\circ {{g}_{2}}\left( -1.09545 \right) \right)=13.1453$
The point is a local minimum
$\Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f\circ {{g}_{2}}\left( 1.09545 \right) \right)=-13.1453$
The point is a local maximum
$\therefore$ There are many local minima and maxima for the given function.

Note: The given question can also be solved plotting a graph for the given function. The absolute maximum of a function is the highest point attained in the graph and absolute minimum is the lowest point attained in the domain of the function.