Question

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, find value of x, y and z.

(a) x=40, y=40, z=40

(b) x=38, y=50, z=32

(c) x=50, y=40, z=30

(d) x=80, y=30, z=50

This question was posed to me in examination.

This interesting question is from Maxima and Minima of Two Variables in chapter Maxima and Minima of Engineering Mathematics

Answer 1

The correct option is (b) x=38, y=50, z=32

The explanation: Now, x + y + z = 120 => z = 120 – x – y

f = xy + yz + zx

f = xy + y(120-x-y) + x(120-x-y) = 120x + 120y – xy – x^2 – y^2

Hence, ^∂f⁄∂x = 120 – y – 2x and ^∂f⁄∂y = 120 – x – 2y

putting ^∂f⁄∂x and ^∂f⁄∂y equals to 0 we get, (x, y)=>(40, 40)

Now at (40,40), r=^∂^2f⁄∂x^2 = -2 < 0, s = ^∂^2f⁄∂x∂y = -1, and t = ^∂^2f⁄∂y^2 = -2

hence, rt – s^2 = 5 > 0

since, r<0 and rt – s^2 > 0 f(x,y) has maixum value at (40,40),

Hence, maximum value of f(40,40) = 120 – 40 – 40 = 40,

Hence, x = y = z = 40.