Question

7) Find three positive numbers whose sum is 100 and whose product is a maximum. Set up a function and use partial derivatives to find the maximum.

Answer 1

we have : ... . * of Substitute X30 in 4 =100-2X Soln; Let X, Y, Z are the integers Then their sum is 100 x+y+=100 and let f(xyz) = xyz 0 can be written as Z=100 -- substitute value o 2 in we get f(xyz) Ryc100-x-4) f(xyz): 100xy - m?y--geye Taking partial derivative write af = (100xy-24-wy?) 9X 222 of = 10oy axy - ye 4 or Taking partial derivative of ③ writ. y. of sy 22 (looxy-a²y_xy) эн 100% - 02-224 sy equate eq" © = 0 looy-axy-y2=0 →y (100-20-4)=0 → Y= 0, y = 100 -2.X substitute eq" =0 100%-7-22y = 0 put y=o x=0 X=100 =100 +4=100.. eq" 1002-12 = 0 .. X=100 x (100-2)=0 substitute y=100-2% in eqh ..100% -2 -22 (100-21) = 0 >100X-2-200% +4-02 = 0 34? 100% =0 (5%-100) = 0

then 2 = 100-2-4 = 100-100- Substitute 2=100 in y = 100 - 204 >= 100 points Thus the ane So critical 10,0), (100,0), (0.100), (100,100) we have following cases @ When (.y)=10,0) ► 2=0, y=0 then 30 f(,y,z) = xyz = 0 When (2,4)= (100,0) 78=100 y=0 then z=100-x-y Z=100 -2-4 Z=100 → Z=100-100-00 (111 So f(2,4,2) = xyz = 0 When x,y) = (0,100) ► 2=0, y=100 then 2 = 100-1-4 = 100-0 -100 =0 So fee,4,2) = beyz = 0 (160,109) When (x,y) 3100 - 200 -2003 300-200 z and f(2,4,2) = xyz = 100*100*100 (100) which is largest among all above possible valises SO are the required integers x= 100 3 y=100 1 Z=100 3 3 s.t. x+y+z=100 and Ryz is maximum.