Find the minimum value of f subject to the given constraint.

Question

f(x,y)=2x2+y2;4x+3y=102

The minimum value of f occurs at

Answer 1

f(x,y) = 2x² + y²; 4x + 3y = 102

 

solve for y from 4x + 3y = 102

3y = 102-4x

y = (102-4x)/3

 

f(x) = 2x² + [(102-4x)/3]²

f(x) = 2x² + (102/3 -4x/3)²

f(x) = 2x² + 10404/9 - (816/9)x + 16x²/9

f(x) = 34x²/9 - (816/9)x + 10404/9

f(x) = 34x²/9 - (816/9)x + 1156

 

This is a parabola opening upwards.  The 

minimum value occurs at the vertex at

V = (-b/2a, f(-b/2a))

 

The x value of the vertex is:

(816/9)/(68/9) = (816/9)(9/68) = 816/68 = 12

 

For x = 12;

y = (102-4(12))/3 = 54/3 = 18

 

The minimum value of f(x, y) is at f(12, 18)

 

f(12, 18) = 2(12)² + 18² = 612

 

The minum value of f(x,y) is 612 at f(12, 18)

Answer 2

f(x,y)=2x^2+y^2

 

If we convert the 2nd equation into a function as well:

g(x,y)=4x+3y-102

 

Use Lagrange multipliers:

fx(x,y)=(lambda)(gx(x,y))

lambda_x = partial derivative of f(x,y) with respect to x=4x

lambda_y=2y

 

set lambda x=lambda y

4x=2y

x=y/2

 

102=4x+3y

102=4(y/2)+3y=2y+3y=5y

y=102/5

 

x=y/2

x=(102/5)/2=51/5

 

Minimum value of f(x,y) occurs at (51/5,102/5)