Question

Find the maximum value of y for which 2x² – 4xy + 3y² – 8x + 8y – 1 = 0

Answer 1

We have 2x² – 4xy + 3y² – 8x + 8y – 1 = 0 ...(i)

⇒ 4x – 4x(dy/dx)  – 4y + 6y

dy/dx – 8 + 8(dy/dx) = 0

⇒ (8 – 4x + 6y)dy/dx = (8 + 4y – 4x)

⇒ dy/dx = (8 + 4y – 4x)/(8 – 4x + 6y)

For max or min, dy/dx = 0

⇒(8 + 4y – 4x)/(8 – 4x + 6y) = 0

⇒ (8 + 4y – 4x) = 0 ⇒ y = x – 2 ...(ii)

On solving (i) and (ii), we get,

2x² – 4x (x – 2) + 3(x – 2)² – 8x + 8(x – 2) = 1

⇒ x² – 4x – 5 = 0

⇒ (x – 5)(x + 1) = 0

⇒ x = –1, 5

when x = 5, then y = 3

when x = –1, then y = –3.

Hence, the maximum value of y is 3.