Finding Maxima and Minima of functions with more than 2 variables

When a function has one variable we know how to find the maxima and minima of the function by differentiating and equating to zero to find the points.

But when a function has more than two variables, we use partial differentiation to find the maxima and minima.

1. f(x,y)=x3+3xy2+2xy$f(x,y)=x^3+3x{y}^2+2xy$ subject to the condition x + y = 4
Sol: The local maximum and minimum of f(x,y) subject to the constraint g(x,y)=0 correspond to the stationary points of L(x,y,λ)=f(x,y)−λ.g(x,y)$L(x,y,\lambda )=f(x,y)-\lambda .g(x,y)$
where λ$\lambda$ is Lagrange multiplier.
We have L(x,y,λ)=x3+3xy2+2xy−λ.(x+y−4)$L(x,y,\lambda )=x^3+3x{y}^2+2xy-\lambda .(x+y-4)$
Now ∂L∂x=3x2+3y2+2y−λ${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }x}=3x^2+3y^2+2y-\lambda }$
λLλy=6xy+2x−λ${\displaystyle \frac{\lambda L}{\lambda y}=6xy+2x-\lambda }$
∂L∂y=6xy+2x−λ${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }y}=6xy+2x-\lambda }$
∂L∂λ=−(x+y−4)${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }\lambda }=-(x+y-4)}$

Note: in doing partial differentiation, except the independent variable everything is considered a constant.  For example, when we do differentiation w.r.t x then, except x all others should be considered constant.
Also ddx(xn)=n.xn−1${\displaystyle \frac{d}{dx}(x^n)=n.x^{n-1}}$

Putting ∂L∂x=∂L∂y=∂L∂λ=0${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }x}=\frac{\mathrm{\partial }L}{\mathrm{\partial }y}=\frac{\mathrm{\partial }L}{\mathrm{\partial }\lambda }=0}$
3x2+3y2+2y=λ$3x^2+3y^2+2y=\lambda$ ------(1)
6xy+2x=λ$6xy+2x=\lambda$ -----(2)
x + y - 4 = 0 ------------(3)
From equations (1) and (2) we get 3x2+3y2+2y=6xy+2x$3x^2+3y^2+2y=6xy+2x$ --------(4)
Putting y = - x + 4 in equation (4)
we get 3x2+3(−x+4)2+2(−x+4)=6x(−x+4)+2x$3x^2+3(-x+4{)}^2+2(-x+4)=6x(-x+4)+2x$
3x2+3(x2−8x+16)−2x+8=−6x2+24x+2x$3x^2+3(x^2-8x+16)-2x+8=-6x^2+24x+2x$
12x2−52x+56=0$12x^2-52x+56=0$
⇒3x2−13x+14=0$\Rightarrow 3x^2-13x+14=0$
⇒(3x−7)(x−2)=0$\Rightarrow (3x-7)(x-2)=0$
⇒x=2,73$\Rightarrow x=2,{\displaystyle \frac{7}{3}}$

For x = 2, from equation (3) we get y = 2 and F(x,y)) = 40
and for x=73$x={\displaystyle \frac{7}{3}}$, y=53$y={\displaystyle \frac{5}{3}}$ and F(x,y) = 392527$39{\displaystyle \frac{25}{27}}$

Alternatively: By substituting y = x - 4 in the equation f(x, y) = x3+3xy2+2xy$x^3+3x{y}^2+2xy$
we get, F(x, 4-x) = x3+3x(4−x)2+2x(4−x)$x^3+3x(4-x{)}^2+2x(4-x)$
F(x) = x3+3x(16−8x+x2)+2x(4−x)$x^3+3x(16-8x+x^2)+2x(4-x)$
F(x) = 4x3−26x2+56x$4x^3-26x^2+56x$
Differentiation F with respect to x we get, F1(x)=12x2−52x+56$F^1(x)=12x^2-52x+56$
Solving like above we get the values of ⇒x=2,73$\Rightarrow x=2,{\displaystyle \frac{7}{3}}$

2. Find the point on the line 3x + 2y = 5 that is closest to the point (3,1)
Sol:  The distance between a general point (x,y) and the point (3, 1) is (x−3)2+(y−1)2−−−−−−−−−−−−−−−√$\sqrt{{(x-3)}^2+{(y-1)}^2}$
We want to find the minimum value of this distance subject to the constraint 3x + 2y = 5.  Infact we have to minimize the square of the distance, and so we minimize f(x, y) = (x−3)2+(y−1)2${(x-3)}^2+{(y-1)}^2$
subject to the given constraint.
L(x,y,λ)=(x−3)2+(y−1)2−λ(3x+2y−5)$L(x,y,\lambda )={\left(x-3\right)}^2+{\left(y-1\right)}^2-\lambda (3x+2y-5)$
∂L∂x=2(x−3)+3λ$\frac{\mathrm{\partial }L}{\mathrm{\partial }x}=2(x-3)+3\lambda$
∂L∂y=2(y−1)+2λ$\frac{\mathrm{\partial }L}{\mathrm{\partial }y}=2(y-1)+2\lambda$
∂L∂λ=−3x−2y+5$\frac{\mathrm{\partial }L}{\mathrm{\partial }\lambda }=-3x-2y+5$

Putting ∂L∂x=∂L∂y=∂L∂λ=0${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }x}=\frac{\mathrm{\partial }L}{\mathrm{\partial }y}=\frac{\mathrm{\partial }L}{\mathrm{\partial }\lambda }=0}$
2(x−3)+3λ=0$2(x-3)+3\lambda =0$ -------(1)
2(y−1)+2λ=0$2(y-1)+2\lambda =0$ ------(2)
3x + 2y = 5 ------------(3)

Multiplying (1) by 2 and (2) by 3 will give
4(x−3)+6λ=0$4(x-3)+6\lambda =0$
6(y−1)+6λ=0$6(y-1)+6\lambda =0$
So 4(x-3) = 6(y-1) ⇒$\Rightarrow$ 2x - 3y = 3 -------(4)
Multiplying equation (3) by 3 and (4) by 2, gives
9x + 6y = 15
4x - 6y = 6
Solving we get x = 2113${\displaystyle \frac{21}{13}}$ and y = 113$\frac{1}{13}$
Thus the point (2113,113)$\left({\displaystyle \frac{21}{13},\frac{1}{13}}\right)$ is on the given line and closest to (3,1)