Discussion
Let C be the curve of intersection of the surfaces \(z^{2} = x^{2} + y^{2}\) and \(4x + z = 7\). If P is a point on C at a minimum distance from the xy-plane, then the distance of P from the origin is
Answer: B
The distance is \(\dfrac{7\sqrt{2}}{5}\).
On the curve, \(z = 7 - 4x\) and \((7 - 4x)^{2} = x^{2} + y^{2}\), so
\[ \begin{aligned} y^{2} &= 15x^{2} - 56x + 49 \\ &= 15\left(x - \tfrac{7}{5}\right)\left(x - \tfrac{7}{3}\right) \ge 0 \end{aligned} \]
which forces \(x \le 7/5\) or \(x \ge 7/3\). The distance from the xy-plane is \(|z| = |7 - 4x|\): on the first branch \(z \ge 7/5\), and on the second \(z \le -7/3\). The minimum is 7/5, at \(P = (7/5, 0, 7/5)\), and
\[ \begin{aligned} |OP| &= \sqrt{\left(\tfrac{7}{5}\right)^{2} + \left(\tfrac{7}{5}\right)^{2}} \\ &= \frac{7\sqrt{2}}{5} \end{aligned} \]
7/5 is the distance from the plane, not from the origin; 14/5 and \(14\sqrt{2}/5\) double the coordinates.
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