Solution Attention! This calculator doesn't check the conditions for applying the method of Lagrange multipliers. Use it at your own risk: the answer may be incorrect.
Rewrite the constraint 4 x 2 + y 2 = 9 4 x^{2} + y^{2} = 9 4 x 2 + y 2 = 9 4 x 2 + y 2 − 9 = 0 4 x^{2} + y^{2} - 9 = 0 4 x 2 + y 2 − 9 = 0
Form the Lagrangian: L ( x , y , λ ) = ( 81 x 2 + y 2 ) + λ ( 4 x 2 + y 2 − 9 ) L{\left(x,y,\lambda \right)} = \left(81 x^{2} + y^{2}\right) + \lambda \left(4 x^{2} + y^{2} - 9\right) L ( x , y , λ ) = ( 81 x 2 + y 2 ) + λ ( 4 x 2 + y 2 − 9 )
Find all the first-order partial derivatives:
∂ ∂ x ( ( 81 x 2 + y 2 ) + λ ( 4 x 2 + y 2 − 9 ) ) = 2 x ( 4 λ + 81 ) \frac{\partial}{\partial x} \left(\left(81 x^{2} + y^{2}\right) + \lambda \left(4 x^{2} + y^{2} - 9\right)\right) = 2 x \left(4 \lambda + 81\right) ∂ x ∂ ( ( 81 x 2 + y 2 ) + λ ( 4 x 2 + y 2 − 9 ) ) = 2 x ( 4 λ + 81 ) partial derivative calculator ).
∂ ∂ y ( ( 81 x 2 + y 2 ) + λ ( 4 x 2 + y 2 − 9 ) ) = 2 y ( λ + 1 ) \frac{\partial}{\partial y} \left(\left(81 x^{2} + y^{2}\right) + \lambda \left(4 x^{2} + y^{2} - 9\right)\right) = 2 y \left(\lambda + 1\right) ∂ y ∂ ( ( 81 x 2 + y 2 ) + λ ( 4 x 2 + y 2 − 9 ) ) = 2 y ( λ + 1 ) partial derivative calculator ).
∂ ∂ λ ( ( 81 x 2 + y 2 ) + λ ( 4 x 2 + y 2 − 9 ) ) = 4 x 2 + y 2 − 9 \frac{\partial}{\partial \lambda} \left(\left(81 x^{2} + y^{2}\right) + \lambda \left(4 x^{2} + y^{2} - 9\right)\right) = 4 x^{2} + y^{2} - 9 ∂ λ ∂ ( ( 81 x 2 + y 2 ) + λ ( 4 x 2 + y 2 − 9 ) ) = 4 x 2 + y 2 − 9 partial derivative calculator ).
Next, solve the system { ∂ L ∂ x = 0 ∂ L ∂ y = 0 ∂ L ∂ λ = 0 \begin{cases} \frac{\partial L}{\partial x} = 0 \\ \frac{\partial L}{\partial y} = 0 \\ \frac{\partial L}{\partial \lambda} = 0 \end{cases} ⎩ ⎨ ⎧ ∂ x ∂ L = 0 ∂ y ∂ L = 0 ∂ λ ∂ L = 0 { 2 x ( 4 λ + 81 ) = 0 2 y ( λ + 1 ) = 0 4 x 2 + y 2 − 9 = 0 . \begin{cases} 2 x \left(4 \lambda + 81\right) = 0 \\ 2 y \left(\lambda + 1\right) = 0 \\ 4 x^{2} + y^{2} - 9 = 0 \end{cases}. ⎩ ⎨ ⎧ 2 x ( 4 λ + 81 ) = 0 2 y ( λ + 1 ) = 0 4 x 2 + y 2 − 9 = 0 .
The system has the following real solutions: ( x , y ) = ( − 3 2 , 0 ) \left(x, y\right) = \left(- \frac{3}{2}, 0\right) ( x , y ) = ( − 2 3 , 0 ) ( x , y ) = ( 0 , − 3 ) \left(x, y\right) = \left(0, -3\right) ( x , y ) = ( 0 , − 3 ) ( x , y ) = ( 0 , 3 ) \left(x, y\right) = \left(0, 3\right) ( x , y ) = ( 0 , 3 ) ( x , y ) = ( 3 2 , 0 ) \left(x, y\right) = \left(\frac{3}{2}, 0\right) ( x , y ) = ( 2 3 , 0 )
f ( − 3 2 , 0 ) = 729 4 f{\left(- \frac{3}{2},0 \right)} = \frac{729}{4} f ( − 2 3 , 0 ) = 4 729
f ( 0 , − 3 ) = 9 f{\left(0,-3 \right)} = 9 f ( 0 , − 3 ) = 9
f ( 0 , 3 ) = 9 f{\left(0,3 \right)} = 9 f ( 0 , 3 ) = 9
f ( 3 2 , 0 ) = 729 4 f{\left(\frac{3}{2},0 \right)} = \frac{729}{4} f ( 2 3 , 0 ) = 4 729
Thus, the minimum value is 9 9 9 729 4 \frac{729}{4} 4 729
Answer Maximum 729 4 = 182.25 \frac{729}{4} = 182.25 4 729 = 182.25 A at ( x , y ) = ( − 3 2 , 0 ) = ( − 1.5 , 0 ) \left(x, y\right) = \left(- \frac{3}{2}, 0\right) = \left(-1.5, 0\right) ( x , y ) = ( − 2 3 , 0 ) = ( − 1.5 , 0 ) ( x , y ) = ( 3 2 , 0 ) = ( 1.5 , 0 ) \left(x, y\right) = \left(\frac{3}{2}, 0\right) = \left(1.5, 0\right) ( x , y ) = ( 2 3 , 0 ) = ( 1.5 , 0 ) A .
Minimum 9 9 9 A at ( x , y ) = ( 0 , − 3 ) \left(x, y\right) = \left(0, -3\right) ( x , y ) = ( 0 , − 3 ) ( x , y ) = ( 0 , 3 ) \left(x, y\right) = \left(0, 3\right) ( x , y ) = ( 0 , 3 ) A .
