69.9k views
0 votes
Kindly solve all the sub-problems of this math

7. Consider the function: \( f(x, y)=4 x^{2}+8 y^{2} \). (a) Find the minimum value of the function subject to \( x+y=0 \). (b) Find the minimum value of the function subject to \( x^{2}+y^{2}=2 \). (

User Tiurin
by
7.9k points

1 Answer

4 votes

Explanation:

(a) To find the minimum value of the function \(f(x, y) = 4x^2 + 8y^2\) subject to the constraint \(x + y = 0\), we can use the method of Lagrange multipliers.

Let's define the Lagrangian function \(L(x, y, \lambda) = f(x, y) - \lambda(g(x, y))\), where \(g(x, y) = x + y\). The parameter \(\lambda\) is the Lagrange multiplier.

Setting up the equations:

\(\frac{\partial L}{\partial x} = 0\)

\(\frac{\partial L}{\partial y} = 0\)

\(g(x, y) = 0\)

Differentiating \(L\) with respect to \(x\) and \(y\):

\(\frac{\partial L}{\partial x} = 8x - \lambda = 0\)

\(\frac{\partial L}{\partial y} = 16y - \lambda = 0\)

\(x + y = 0\)

From the first equation, we have \(8x = \lambda\), and from the second equation, we have \(16y = \lambda\). Equating these two expressions, we get \(8x = 16y\), which simplifies to \(x = 2y\).

Substituting this into the constraint equation \(x + y = 0\), we get \(2y + y = 0\), which yields \(y = 0\). From here, we can find \(x = 0\).

Therefore, the only critical point satisfying the constraint is \((x, y) = (0, 0)\).

To determine if this critical point is a minimum or maximum, we can evaluate the function \(f(x, y) = 4x^2 + 8y^2\) at this point:

\(f(0, 0) = 4(0)^2 + 8(0)^2 = 0\)

The minimum value of the function subject to the constraint \(x + y = 0\) is \(f(0, 0) = 0\).

(b) Without the specific constraint mentioned in part (b), we cannot determine the minimum value of the function. The function \(f(x, y) = 4x^2 + 8y^2\) does not have a global minimum unless there is a specific constraint or boundary condition specified.

User DanHickstein
by
8.6k points

No related questions found