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Let x, y, z be real numbers such that 4x^2 + y^2 + 16z^2 = 1. Find the maximum value of 7x + 2y + 8z.

Apr 14, 2020

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By Lagrange multipliers,

$$\begin{cases} \nabla_{x,y,z}{\left(4x^2 + y^2 + 16z^2 - 1\right)} = \lambda \nabla_{x,y,z}\left(7x + 2y + 8z\right)\\ 4x^2 + y^2 + 16z^2 = 1 \end{cases}$$

$$\begin{cases} x = \dfrac78 \lambda\\ y = \lambda\\ z = \dfrac14\lambda\\ 4x^2 + y^2 + 16z^2 = 1 \end{cases}$$

Substitute the first three equations to the fourth one,

$$\dfrac{49}{16}\lambda^2 + \lambda^2 + \lambda^2 = 1\\ \lambda^2 = \dfrac{16}{81}\\ \lambda = \pm \dfrac 49$$

Optimum occurs when $$(x,y,z) = \left(\dfrac7{18}, \dfrac49, \dfrac19\right)$$ or $$(x,y,z) = \left(-\dfrac7{18}, -\dfrac49, -\dfrac19\right)$$

Substitute these values to find that the maximum value of 7x + 2y + 8z occurs when $$(x,y,z) = \left(\dfrac7{18}, \dfrac49, \dfrac19\right)$$, and the value is $$\dfrac92$$.

Apr 14, 2020