Answer:
Minimum value: 0 (at x = 0, y = 0)
Maximum value: 3^11 * 3/2 (at x = 81/4, y = 27/2)
Explanation:
To find the extreme values of the given function, we need to use the method of Lagrange multipliers. Let's define the following three functions:
f(x,y) = x^2 y^2 (12 - 3x - 4y) (the given function)
g1(x,y) = 2xy^2 - 6x^2y (partial derivative of f with respect to x)
g2(x,y) = 2x^2y - 4xy^2 (partial derivative of f with respect to y)
Now, we need to solve the system of equations:
g1(x,y) = λ dg1/dx
g2(x,y) = λ dg2/dy
where λ is the Lagrange multiplier. Taking the partial derivatives of g1 and g2 with respect to x and y, respectively, we get:
dg1/dx = 2y^2 - 12xy
dg2/dy = 2x^2 - 8xy
So, the system of equations becomes:
2xy^2 - 6x^2y = λ (2y^2 - 12xy)
2x^2y - 4xy^2 = λ (2x^2 - 8xy)
Dividing the second equation by the first, we get:
x/y = 3/2
Substituting this into the first equation, we get:
4x^3 - 81x^2 = 0
Solving for x, we get x = 0 or x = 81/4. Substituting these values back into the equation x/y = 3/2, we get y = 0 or y = 54/4 = 27/2.
Now, we need to check the extreme values of the function at the critical points (0,0) and (81/4, 27/2), as well as at the endpoints of the domain.
At (0,0), the value of the function is 0.
At (81/4, 27/2), the value of the function is:
(81/4)^2 (27/2)^2 (12 - 3(81/4) - 4(27/2)) = 81^2 (27/2) (3/4) = 3^11 * 3/2
At the endpoints of the domain, we need to check the values of the function at the boundary lines x = 0, x = 12/3 = 4, y = 0, and y = 12/4 = 3. The values of the function at these lines are:
x = 0: 0
x = 4: 4^2 y^2 (12 - 3(4) - 4y) = -256y^2
y = 0: 0
y = 3: x^2 3^2 (12 - 3x - 4(3)) = 27x^2 (12 - 3x - 12) = -81x^3 + 972x^2
So, the extreme values of the function are:
Minimum value: 0 (at x = 0, y = 0)
Maximum value: 3^11 * 3/2 (at x = 81/4, y = 27/2)