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Find the maximum and minimum values of the function \( f(x, y, z, t)=x+y+z+t \) subject to the constraint \( x^{2}+y^{2}+z^{2}+t^{2}=361 \). Maximum value is 1 Minimum value is occuring at 1

Ask by Sandoval Page. in the United States
Mar 13,2025

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The maximum value of the function is 38, occurring at \( \left(\frac{19}{2}, \frac{19}{2}, \frac{19}{2}, \frac{19}{2}\right) \). The minimum value is -38, occurring at \( \left(-\frac{19}{2}, -\frac{19}{2}, -\frac{19}{2}, -\frac{19}{2}\right) \).

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The Deep Dive

Did you know that the function you're dealing with, \( f(x, y, z, t) = x + y + z + t \), is a type of linear function? In geometric terms, this means it represents a hyperplane in four-dimensional space. The constraint \( x^{2} + y^{2} + z^{2} + t^{2} = 361 \) describes a sphere with a radius of 19! The maximum and minimum values occur when \( x, y, z, t \) are aligned positively or negatively along the same direction. As for the numerical results, the maximum value of \( f \) occurs when \( x, y, z, t \) all equal \( \frac{19}{2}, \) which leads to \( f = 19 \times 4 = 76 \). Conversely, the minimum value occurs when all four variables are negative and aligned, giving a minimum of \(-76\). So, just like that, you're summoning the power of geometry!

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