Question

7. [Global Optimization on Domains] [P] Find the absolute maximum and minimum values of the function on the described domain. When checking for extrema on any boundaries, you may choose whether you want to use a parametrization or Lagrange Multipliers; some problems might be easiest when using a combination. (a)f(x,y)=x 2 −y 2 −2xy−2x, on the triangular region whose vertices are(1,0),(0,1), and(−1,0). (b)g(x,y)=x+xy−2y 2, on the domainy≥0,y≤ x​,x≤1. (c)h(x,y)=e xy, on the disk( 2x​ ) 2 +y 2 ≤1. t† If you use your answers to part (a) and part (b) to justify why this is a maximum, you don't have to do the second rivative test. (d)f(x,y)=xy, on the part of the curve2x 3 +y 3 =16which is in the first quadrant (including endpoints). (e)g(x,y)=x 2 +3y, on the line segment from(−2,2)to(2,−2). (f)h(x,y,z)=x+2y+3z, on the unit sphere.

Answer 1

(a) The Absolute minimum value is -3/2 and maximum value is 3.

(b) We have global minima at x = 4/9 and global minimum value at x = −4/27.

(c) Global maximum value of function is 2.72 and minimum value of function is 0.37.

a) Given function as:

f(x,y)=x2-y2-2xy-2x

The triangular region described by the vertices (1,0),(0,1) and (-1,0) can be described by the straight lines :

a) Straight line joining (1,0) and (0,1) i.e y= 1-x

b) Straight line joining (0,1) and (-1,0) i.e y= 1+x

c) Straight line joining (-1,0) and (1,0) i.e the x axis where y=0

When y=0, we have, x^2-2x

Now, df/dx=0 for maxima / minima

2x-2=0

X=1

Also, d^(2)f/dx^(2)=2>0 and df/dx<0 for x<1

So, we have minimum value = 12−2×1=−1

and maximum value = (−1)2−2×−1=3

When y=1+x, we have ,

f(x,y)=x^2−(1+x)^2−2x(1+x)−2x

df/dx=2x−2(1+x)−2x−2(1+x)−2

=−6−4x<0, for  x>−3/2

Hence f(x,y) is always decreasing function in the interval x=[-1,0] , with minimum

=f(0,1)

=02−12−2×0×1−2×0

=−1

Maximum value = f(-1,0)

=(−1)2−02−2×−1×0−2×−1

=1−0+2

=3

When y= 1-x, we have,

f(x,y)=x^2−(1−x)^2−2x(1−x)−2x

df/dx=2x+2(1−x)+2x−2(1−x)−2

=4x−2>0,for x>1/2

Hence at x=1/2,we have df/dx=0 which is global minima point since

d2(f)/dx2=4>0

So, minimum value for y = 1- x

=f(1/2,1/2)

=1/22−1/22−2×1/2×1/2−2×1/2

=−1/2−1=−3/2

Maximum value in the interval [ 0 1] shall be

=max(f(0,1).f(1,0))

=max(−1,−1)=−1

So, considering all points on this triangle , we have absolute maximum

= max(f(x,y) for y=0, f(x,y) for y=1+x, f(x,y) for y=1-x)

=max(3,3,−1)

=3

Absolute minimum considering all points on the triangle

= min(f(x,y) for y=0, f(x,y) for y=1+x, f(x,y) for y=1-x)

=min(−1,−1,−1.5)

=−3/2

b) Given g(x,y)=x+xy−2y2

When y=x0.5, we have,

g(x,y)=x+x^1.5−2x

=x^1.5−x

For the domain x = [0 , 1], we have ,

dg/dx=1.5×x^0.5−1=0

x=4/9

d^2(g)/d^x2=0.75x0.5>0 for x=4/9

Hence, we have global minima at x = 4/9 and global minimum value

=(4/9)1.5−49

=8/27−4/9

=−4/27

Global maximum = max(g(0,0),g(1,1))

=max(0,0)

=0

c) Given h(x,y)=exp⁡(xy)

When (x/2)^{2}+y^{2}=1, we have

y=(1−x^{2}/4)^0.5

So, h(x,y) = $\text{exp}\left(x(1-\frac{x^{2}}{4})^{0.5}\right)$

Now,dh/dx=$\text{exp}\left(x(1-\frac{x^{2}}{4})^{0.5}\right) \times ((1-\frac{x^2}{4})^{0.5}+0.5x\times\frac{-x/2}{(1-x^2/4)^{0.5}}$

For maxima / minima , we mush have ,

dh/dx=0

$(1-\frac{x^2}{4})^{0.5}+0.5x\times\frac{-x/2}{(1-x^2/4)^{0.5}}$5=0

$(1-x^2/4)^{0.5}-0.25 \frac{x^2}{(1-x^2/4)^{0.5}}$=0

$1-x^2/4-0.25x^2$=0

1−x^2/2=0

x = 1.41,−1.41

In the range [-1.41,1.41], we always have dh/dx>0

At x=20.5and x=−20.5, we have

y=(1−x2/4) 0.5

=(1−24)0.5

=120.5,−120.5

Global maximum value of function

=f(20.5,1/20.5)

=exp(1)

=2.72

Global minimum value of function

=f(20.5,−1/20.5)

=exp(−1)

=(0.37)

To learn more about Global maximum and minimum value link is here

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