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zmudz

  • one year ago

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

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  1. myininaya
    • one year ago
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    You could use lagrange multpliers.

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

  3. Zarkon
    • one year ago
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    bookmark

  4. beginnersmind
    • one year ago
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    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.

  5. thomas5267
    • one year ago
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    Cheaty generalised mean inequality: \[ x_1=\frac{9x}{81}\\ x_2=\frac{12y}{144}\\ x_3=\frac{8z}{64}\\ \sqrt{\frac{1}{289}(81x_1^2+144x_2^2+64x_3^2)}=\frac{1}{17}\sqrt{x^2+y^2+z^2}=\frac{1}{17}\geq\frac{81x_1+144x_2+64x_3}{289}\\ 17\geq 81x_1+144x_2+64x_3=9x+12y+8z\\ 17= 81x_1+144x_2+64x_3=9x+12y+8z\text{ iff }x_1=x_2=x_3 \]

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