Need Help?

Subscribe to Calculus A

Subscribe
  • Notes
  • Comments & Questions

Given $$\longrightarrow X Y=100 \quad x>0 \quad Y>0$$

$$f=x+y \quad$$ but $$\quad y=\frac{100}{x}$$

$$f=x+\frac{100}{x} \longrightarrow f^{\prime}(x)=1-\frac{100}{x^{2}} \rightarrow f^{\prime}(x)=\frac{x^{2}-100}{x^{2}}$$

$$f^{\prime}(x)=0 \longrightarrow \frac{x^{2}-100}{x^{2}}=0 \rightarrow x^{2}-100=0 \rightarrow(x+10)(x-10)=0$$

$$x=10, x=-10$$ rejected

$$x=10$$

Applying Second Derivative Test:

$$f^{\prime \prime}(x)=\frac{-100 (2 x)(-1)}{x^{4}}=\frac{200}{x^{3}}$$

$$f^{\prime \prime}(10)=\frac{200}{1000}>0$$

\ F has Iocal minimum at $$x=10$$

$$y=\frac{100}{x}=\frac{100}{10}=10$$

$$f=x+y=10+10=20$$

$$x y=10(10)=100$$

No comments yet

Join the conversation

Join Notatee Today!