Divide 120 into three parts so that the sum of their products taken two at a time is maximum. If x, y, z are two parts, then the value of x, y and z will be
Answers
Answer:
Step-by-step explanation:
Let x+y+z =120 and f(x,y,z) = xy + yz + zx. Then F(x,y) = f(x,y,120-(x+y)) = xy+(x+y)z = xy+120(x+y)-(x+y)^2.
We have (F_x) = y+120 - 2(x+y) and
(F_y) = x +120 - 2(x+y) and equating both to 0 gives x = y = 40 ==> z= 120-(x+y) = 40 also. Hence f(x,y,z) = 3×40×40 = 4800. Now note that by taking x= y = 0 and z = 120, the given condition holds but f(0,0,120) = 0 < 4800. Hence the value 4800 is not the minimum value and hence must be the maximum value. Hence the answer is 120
40 , 40 , 40 are three parts of 120 such that the sum of there product taken two at a time is maximum
Step-by-step explanation:
Let say three part are
x , y , z
z = (120 - x - y)
Sum of three products = xy + yz + zx
=> S = xy + y (120 - x - y) + x(120 - x - y)
=> S = 120x + 120y - xy - x² - y²
∂/S/∂x = 120 -y - 2x
∂/S/∂y = 120 -x - 2y
120 -y - 2x = 120 -x - 2y
=> y = x
S = 120x + 120x - x² - x² - x²
=> S = 240x - 3x²
dS/dx = 240 - 6x
=> 240 - 6x = 0
=> x = 40
d²S/dx² = - 6 ( -ve hence maximum value)
=> x = 40
y = 40
z = 120 -x - y = 40
40 , 40 , 40 are three parts of 120 such that the sum of there product taken two at a time is maximum
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