Lesson Solving non-linear minimax problems in 3D space
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<H2>Solving non-linear minimax problems in 3D space</H2> <H3>Problem 1</H3>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. <B>Solution</B> <pre> By analogy with the well known AM-GM inequality ("Arithmetic Mean - Geometric Mean inequality") for two variables "a" and "b" ab <= {{{((a+b)/2)^2}}}, (1) there is AM-GM inequality for three variables "a", "b" and "c" abc <= {{{((a + b + c)/3)^3}}}. (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) <= {{{((x-3)+(y-5)+(z-2))/3)^3}}} = {{{((x+y+z - 3-5-2))/3)^3}}} = {{{((16-10)/3)^3}}} = {{{6/3)^3}}} = {{{2^3}}} = 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. <U>ANSWER</U> </pre> <H3>Problem 2</H3>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. <B>Solution</B> <pre> By analogy with the well known AM-GM inequality ("Arithmetic Mean - Geometric Mean inequality") for two variables "a" and "b" ab <= {{{((a+b)/2)^2}}}, (1) there is AM-GM inequality for three variables "a", "b" and "c" abc <= {{{((a + b + c)/3)^3}}}. (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 <= {{{((x^2 + y^2 + z^2)/3)^3}}} = {{{(196/3)^3}}} = {{{196^3/3^3}}}. Thus the maximum value of x^2*y^2*z^2, under the constraint x^2+y^2+z^2 = 196 is {{{(196/3)^3}}} = {{{7529536/27}}}. <U>ANSWER</U> It is achieved when x^2 = y^2 = z^2 = {{{196/3}}}, i.e. x = y = z = +/- {{{sqrt(196/3)}}} = 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. </pre> My other additional lessons on Miscellaneous word problems in this site are - <A HREF=https://www.algebra.com/algebra/homework/word/misc/I-do-not-have-enough-savings-now.lesson>I do not have enough savings now</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/In-a-jar-all-but-6-are-red-marbles.lesson>In a jar, all but 6 are red marbles</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/How-many-boys-and-how-many-girls-are-there-in-a-family.lesson>How many boys and how many girls are there in a family ?</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/What-is--the-last-digit-of-the-number-a%5En-.lesson>What is the last digit of the number a^n ?</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/Find-the-last-three-digits-of-these-numbers-.lesson>Find the last three digits of these numbers</A> - 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