Lesson Solving non-linear minimax problems in 3D space

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Solving non-linear minimax problems in 3D space


Problem 1

If   x+y+z=16,  then find the maximum value of  (x-3)(y-5)(z-2),  given that  (x-3) > 0,  (y-5) > 0,  (z-2) > 0.

Solution 

By analogy with the well known  AM-GM inequality ("Arithmetic Mean - Geometric Mean inequality") for two variables "a" and "b"


    ab <= %28%28a%2Bb%29%2F2%29%5E2,         (1)


there is AM-GM inequality for three variables "a", "b" and "c"


    abc <= %28%28a+%2B+b+%2B+c%29%2F3%29%5E3.      (2)


Inequalities (1) and (2) are valid for any two and three variables, respectively, that are real non-negative numbers.


Apply inequality (2), taking  

    a = x-3,  b = y-5,  c = z-2.


You will get

    (x-3)*(y-5)*(z-2) <= %28%28x-3%29%2B%28y-5%29%2B%28z-2%29%29%2F3%29%5E3 = %28%28x%2By%2Bz+-+3-5-2%29%29%2F3%29%5E3 = %28%2816-10%29%2F3%29%5E3 = 6%2F3%29%5E3 = 2%5E3 = 8.


Thus for any 3 values of x, y and z, restricted by the equality

    x + y + z = 16   and  inequalities  x >= 3, y >= 5  and  z >= 2,

the inequality  

    (x-3)*(y-5)*(z-2) <= 8 

is held.


From the other side, at x= 5, y= 7  and z= 4 we have

    (x-3)*(y-5)*(z-2) = (5-3)*(7-5)*(4-2) = 2*2*2 = 8.

and the values of x, y and z satisfy all needed restrictions.



Thus the maximum value of (x-3)*(y-5)*(z-2), where  x, y and z are restricted by 

    x + y + z = 16,    x >= 3, y >= 5  and  z >= 2

is 8.        ANSWER

Problem 2

Find the maximum values of the function   f(x,y,z) = x^2y^2z^2   subject to the constraint   x^2+y^2+z^2 = 196.

Solution 

By analogy with the well known  AM-GM inequality ("Arithmetic Mean - Geometric Mean inequality") for two variables "a" and "b"

    ab <= %28%28a%2Bb%29%2F2%29%5E2,         (1)


there is AM-GM inequality for three variables "a", "b" and "c"

    abc <= %28%28a+%2B+b+%2B+c%29%2F3%29%5E3.      (2)


    Inequalities (1) and (2) are valid for any two and three variables, respectively, that are real non-negative numbers.


    In inequalities (1) and (2), equalities are achieved if and only if  a = b  (for (1))  or  a = b = c (for (2)).



Apply inequality (2), taking  

    a = x^2,  b = y^2,  c = z^2.


You will get

    x^2*y^2*z^2 <= %28%28x%5E2+%2B+y%5E2+%2B+z%5E2%29%2F3%29%5E3 = %28196%2F3%29%5E3 = 196%5E3%2F3%5E3.


Thus the maximum value of  x^2*y^2*z^2,  under the constraint  x^2+y^2+z^2 = 196  is  %28196%2F3%29%5E3 = 7529536%2F27.    ANSWER


It is achieved when  x^2 = y^2 = z^2 = 196%2F3,  i.e.  x = y = z = +/- sqrt%28196%2F3%29 = 8.082904  (rounded).


In all, there are 8 points on the 3D sphere surface  x^2 + y^ + z^2 = 196,
where the maximum value of x^2*y^2*z^2 is achieved - one such point in each octant.


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