Question

1. Find the maxima of the following functions. (a) f(x)-2-4. )2 (c) f (z,y)2+3. (d) f (x,y) = xy-x2-y2 + 9y.

Answer 1

(a) y = f(x) = -x² - x⁴

For an extrema, f'(x) = dy/dx = -2x - 4x³ = 0

2x + 4x³ = 0

2x(1 + 2x²) = 0

Either x = 0 or (1 + 2x²) = 0, which means 2x² = -1, or x² = -1/2 which is inadmissible.

The only critical value of x is x = 0, for which f(x) = 0.

(b) y = f(x) = -x² + x

For an extrema, f'(x) = dy/dx = -2x + 1 = 0

2x = 1

x = 1/2 (critical value of x)

f(x) = -(1/4) + (1/2) = 1/4

(c) z = f(x, y) = -x² - y⁴ + 3

For an extrema, z_x = 0 and z_y = 0

z_x = $\partial$ z/ $\partial$ x = -2x = 0

x = 0

z_y = $\partial$ z/ $\partial$ y = -4y³ = 0

y³ = 0

y = 0

Therefore, critical x = 0 and critical y = 0, for which f(x, y) = 3

(d) z = f(x, y) = xy - x² - y² + 9y

For an extrema, z_x = 0 and z_y = 0

z_x = $\partial$ z/ $\partial$ x = y - 2x = 0

y = 2x

z_y = $\partial$ z/ $\partial$ y = x - 2y + 9 = 0

x - 2y = -9

x - 2(2x) = -9

x - 4x = -9

-3x = -9

x = 3

y = 2 x 3 = 6

Therefore, critical x = 3 and critical y = 6, for which f(x, y) = (3 x 6) - (3 x 3) - (6 x 6) + (9 x 6) = 18 - 9 - 36 + 54 = 27