10. A hydrogen-powered car manufacturer can produce \( N \) number of cars according to the function, \[ N(x, y)=(x+2)^{2}+(y-1)^{2} \] where \( x \) and \( y \), are factory parameters. The factory operation is restricted by the following equation, \[ x^{2}+y^{2}=125 \] Use Lagrange multipliers to find the maximum and minimum values of \( N \) subject to the given constraint, if such values exist.

To apply Lagrange multipliers, we need to find the critical points of the function \( N(x, y) = (x + 2)^2 + (y - 1)^2 \) subject to the constraint \( g(x, y) = x^2 + y^2 - 125 = 0 \). First, we compute the gradients: 1. Gradient of \( N \): \[ \nabla N = \left( \frac{\partial N}{\partial x}, \frac{\partial N}{\partial y} \right) = \left( 2(x + 2), 2(y - 1) \right) \] 2. Gradient of \( g \): \[ \nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = \left( 2x, 2y \right) \] Setting up the equations for Lagrange multipliers: \[ \nabla N = \lambda \nabla g \] This gives us the system of equations: \[ 2(x + 2) = \lambda (2x) \quad (1) \] \[ 2(y - 1) = \lambda (2y) \quad (2) \] Along with the constraint: \[ x^2 + y^2 = 125 \quad (3) \] From equation (1): \[ x + 2 = \lambda x \implies x(1 - \lambda) = -2 \implies x = \frac{-2}{1 - \lambda} \quad (if \, x \neq 0) \quad (4) \] From equation (2): \[ y - 1 = \lambda y \implies y(1 - \lambda) = 1 \implies y = \frac{1}{1 - \lambda} \quad (if \, y \neq 0) \quad (5) \] Substituting (4) and (5) into the constraint (3): \[ \left( \frac{-2}{1 - \lambda} \right)^2 + \left( \frac{1}{1 - \lambda} \right)^2 = 125 \] \[ \frac{4 + 1}{(1 - \lambda)^2} = 125 \] \[ \frac{5}{(1 - \lambda)^2} = 125 \implies (1 - \lambda)^2 = \frac{5}{125} = \frac{1}{25} \] Thus, we have: \[ 1 - \lambda = \pm \frac{1}{5} \] This leads to two solutions for \( \lambda \): 1. \( 1 - \lambda = \frac{1}{5} \implies \lambda = \frac{4}{5} \) 2. \( 1 - \lambda = -\frac{1}{5} \implies \lambda = \frac{6}{5} \) Now solving for \( \lambda = \frac{4}{5} \): Substituting back into equations (4) and (5): \[ x = \frac{-2}{\frac{1}{5}} = -10, \quad y = \frac{1}{\frac{1}{5}} = 5 \] From (3): \[ (-10)^2 + 5^2 = 100 + 25 = 125 \quad (valid) \] Now solving for \( \lambda = \frac{6}{5} \): \[ x = \frac{-2}{-\frac{1}{5}} = 10, \quad y = \frac{1}{-\frac{1}{5}} = -5 \] From (3): \[ (10)^2 + (-5)^2 = 100 + 25 = 125 \quad (valid) \] Now we have two points: \( (-10, 5) \) and \( (10, -5) \). Calculating \( N \) at \( (-10, 5) \): \[ N(-10, 5) = (-10 + 2)^2 + (5 - 1)^2 = (-8)^2 + (4)^2 = 64 + 16 = 80 \] Calculating \( N \) at \( (10, -5) \): \[ N(10, -5) = (10 + 2)^2 + (-5 - 1)^2 = (12)^2 + (-6)^2 = 144 + 36 = 180 \] Thus, the maximum