Question

Need help with Cal 1

Answer 1

both?

Answer 2

\[xy=242\]
\[y=\frac{242}{x}\]
\[2x+y=P\]
\[P(x)=2x+\frac{242}{x}\]
\[P'(x)=2-\frac{242}{x^2}\]
\[2-\frac{242}{x^2}=0\]
\[x^2-242=0\]
\[x^2=242\]
\[x=\sqrt{242}=11\sqrt{2}\]

Answer 3

hmm i must have messed up somewhere but i don't see where. answer should be 22 by 11

Answer 4

yes sorry at work just got back

Answer 5

thank you so much