Answer:
f(x,y) = is maximum at x = 2 and y = 2 and f(2,2) =
Step-by-step explanation:
Since f(x,y) = and x³ + y³ = 16, Ф(x,y) = x³ + y³ - 16
df/dx = y, df/dy = x
, dФ/dx = 3x² and dФ/dy = 3y²
From the method of Lagrange multipliers,
df/dx = λdΦ/dx and df/dy = λdΦ/dy
y = 3λx² (1) and x
= 3λy² (2)
multiplying (1) by x and (2) by y, we have
xy = 3λx³ (4) and xy
= 3λy³ (5)
So, 3λx³ = 3λy³
⇒ x = y
Substituting x = y into the constraint equation, we have
x³ + y³ = 16
x³ + x³ = 16
2x³ = 16
x³ = 16/2
x³ = 8
x = ∛8
x = 2 ⇒ y = 2, since x = y
So, f(x,y) = f(2,2) = =
We need to determine if this is a maximum or minimum point by considering other points that fit into the constraint equation.
Since x³ + y³ = 16 when x = 0, y is maximum when y = 0, x = maximum
So, 0³ + y³ = 16
y³ = 16
y = ∛16
Also, when y = 0, x = maximum
So, x³ + 0³ = 16
x³ = 16
x = ∛16
and f(0,∛16) = .
Also, f(∛16, 0) = .
Since f(0,∛16) = f(∛16, 0) = 1 < f(2,2) =
f(2,2) is a maximum point