Answer:
A correspoding
Step-by-step explanation:
Use both!
You want to minimize <em>P</em>, so differentiate <em>P</em> with respect to <em>x</em> and set the derivative equal to 0 and solve for any critical points.
<em>P</em> = 8/<em>x</em> + 2<em>x</em>
d<em>P</em>/d<em>x</em> = -8/<em>x</em>² + 2 = 0
8/<em>x</em>² = 2
<em>x</em>² = 8/2 = 4
<em>x</em> = ± √4 = ± 2
You can then use the second derivative to determine the concavity of <em>P</em>, and its sign at a given critical point decides whether it is a minimum or a maximum.
We have
d²<em>P</em>/d<em>x</em>² = 16/<em>x</em>³
When <em>x</em> = -2, the second derivative is negative, which means there's a relative maximum here.
When <em>x</em> = 2, the second derivative is positive, which means there's a relative minimum here.
So, <em>P</em> has a relative maximum value of 8/(-2) + 2(-2) = -8 when <em>x</em> = -2.
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