what is the shortest distance from the surface ( xy + 15x+z^{2}=201 ) to the origin? distance =

Explanation:

Step1: Set up the distance formula and the constraint

The distance from a point ((x,y,z)) to the origin is (d=\sqrt{x^{2}+y^{2}+z^{2}}). To simplify calculations, we can minimize (f(x,y,z)=x^{2}+y^{2}+z^{2}) subject to the constraint (g(x,y,z)=xy + 15x+z^{2}-201=0).

Step2: Use Lagrange multipliers

The Lagrange function is (L(x,y,z,\lambda)=x^{2}+y^{2}+z^{2}-\lambda(xy + 15x+z^{2}-201)). Take partial derivatives: (\frac{\partial L}{\partial x}=2x-\lambda(y + 15)=0) (1) (\frac{\partial L}{\partial y}=2y-\lambda x=0) (2) (\frac{\partial L}{\partial z}=2z-2\lambda z = 0) (3) (\frac{\partial L}{\partial\lambda}=-(xy + 15x+z^{2}-201)=0) (4) From (3): (2z-2\lambda z=(2 - 2\lambda)z = 0). So either (z = 0) or (\lambda=1). If (z = 0): From (1): (2x-\lambda(y + 15)=0\Rightarrow\lambda=\frac{2x}{y + 15}) From (2): (2y-\lambda x=0\Rightarrow\lambda=\frac{2y}{x}) Then (\frac{2x}{y + 15}=\frac{2y}{x}\Rightarrow x^{2}=y(y + 15)) Substitute (z = 0) into (4): (xy+15x=201\Rightarrow x(y + 15)=201) Since (x^{2}=y(y + 15)) and (x(y + 15)=201), then (y=\frac{x^{2}}{y + 15}). Let (t=y + 15), then (x t=201) and (y=t - 15), (x^{2}=(t - 15)t). (x^{2}=t^{2}-15t) and (x=\frac{201}{t}), so (\left(\frac{201}{t}\right)^{2}=t^{2}-15t) (201^{2}=t^{4}-15t^{3}) Another case: if (\lambda = 1) From (1): (2x-(y + 15)=0\Rightarrow y=2x-15) From (2): (2y-x=0). Substitute (y = 2x-15) into (2y-x=0) (2(2x-15)-x=0\Rightarrow4x-30 - x=0\Rightarrow3x=30\Rightarrow x = 10) If (x = 10), then (y=5) Substitute (x = 10,y = 5) into (4): (10\times5+15\times10+z^{2}=201) (50+150+z^{2}=201\Rightarrow z^{2}=1)

Step3: Calculate the distance

If (x = 10,y = 5,z=\pm1) (d=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{10^{2}+5^{2}+1^{2}}=\sqrt{100 + 25+1}=\sqrt{126}=3\sqrt{14})

Answer:

(3\sqrt{14})