Question

Let \(x\), \(y\), and \(z\) be real numbers such that \(x^2 + y^2 + z^2 = 1.\) Find the maximum value of \(9x+12y+8z.\)

Answer 1

You could use lagrange multpliers.

Answer 2

\[\text{ Let } g(x,y,z)=x^2+y^2+z^2 \\ \text{ let } f(x,y,z)=9x+12y+8z \\ \text{ solve the following system } \\ f_x=\lambda g_x \\ f_y= \lambda g_y \\ f_z=\lambda g_z \\ x^2+y^2+z^2=1\]

Answer 3

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Answer 4

Using @myininaya 's notation, g is a sphere of radius 1 centered on (0,0,0) and the level sets of f are parallel planes. We're looking for the plane that's a distance of 1 from the point (0,0,0). There should be 2 of these.

We know that the radius vector is normal to plane in the tangent point. So, n = (9,12,8) or n = (-9,-12,-8). Normalize and plug into f.

Answer 5

Cheaty generalised mean inequality:
\[
x_1=\frac{9x}{81}\\
x_2=\frac{12y}{144}\\
x_3=\frac{8z}{64}\\
\sqrt{\frac{1}{289}(81x_1^2+144x_2^2+64x_3^2)}=\frac{1}{17}\sqrt{x^2+y^2+z^2}=\frac{1}{17}\geq\frac{81x_1+144x_2+64x_3}{289}\\
17\geq 81x_1+144x_2+64x_3=9x+12y+8z\\
17= 81x_1+144x_2+64x_3=9x+12y+8z\text{ iff }x_1=x_2=x_3
\]