Question

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y)= e^xy; X^3+y^3=16

Answer 1

Answer:

f(x,y) = $e^{xy}$ is maximum at x = 2 and y = 2 and f(2,2) = $e^{4}$

Step-by-step explanation:

Since f(x,y) = $e^{xy}$ and x³ + y³ = 16, Ф(x,y) = x³ + y³ - 16

df/dx = y$e^{xy}$, df/dy = x$e^{xy}$, dФ/dx = 3x² and dФ/dy = 3y²

From the method of Lagrange multipliers,

df/dx = λdΦ/dx and df/dy = λdΦ/dy

y$e^{xy}$ = 3λx² (1) and  x$e^{xy}$ = 3λy² (2)

multiplying (1) by x and (2) by y, we have

xy$e^{xy}$ = 3λx³ (4) and  xy$e^{xy}$ = 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) = $e^{2 X2}$ = $e^{4}$

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) = $e^{0X\sqrt[3]{16} } = e^{0} = 1$.

Also, f(∛16, 0) = $e^{\sqrt[3]{16}X0 } = e^{0} = 1$.

Since f(0,∛16) = f(∛16, 0) = 1 < f(2,2) = $e^{4}$

f(2,2) is a maximum point