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June 7th, 2015, 06:43 PM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  find minimum with log
Hi, I need help with this two problems. Please someone show me the way to solve them! Thanx

June 7th, 2015, 08:37 PM  #2 
Senior Member Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus 
You should use derivatives to find the minimum value, i think this question is of calculus, differentiate the function with respect to x and then solve it for the point where it is becoming negative.Find y in terms of x, then find dy/dx and then solve the inequality dy/dx<0, find x and then find the value of y at this x.

June 7th, 2015, 10:09 PM  #3 
Math Team Joined: Nov 2014 From: Australia Posts: 688 Thanks: 243 
I think I have a solution for the second one without calculus. From the restriction that $xy = 8$, we may obtain an upper bound on $x$ and $y$. That is, $2\leq x\leq 16$ and $\dfrac{1}{2}\leq y\leq 4$. Taking the log base 2 of both sides of the equation $xy = 8$, we obtain $$\log_2xy = \log_2x + \log_2y = 3$$ So that $$\log_2y = 3  \log_2x$$ From this, we obtain $$\begin{align*} &\,\,\,\,\,\,\,\,\log_2\,x\cdot\log_2\,y \\\\ &= \log_2x(3  \log_2x) \\\\ &= 3\log_2x  (\log_2x)^2 \\\\ &= \left((\log_2x)^2  3\log_2\,x + \dfrac{9}{4}\right) + \dfrac{9}{4} \\\\ &= \dfrac{9}{4}  \left(\log_2x  \dfrac{3}{2}\right)^2 \end{align*}$$ This expression is at its maximum when the second term is $0$. This occurs when $\log_2x = \dfrac{3}{2}$. This implies that $x = 2^{2/3}$, which is within the permitted range. Thus the maximum is $\dfrac{9}{4}$. The minimum occurs when the second term is as large as possible. We try the extreme values of $x$, namely $2$ and $16$. If $x = 2$, $\log_2x\cdot\log_2y = 2$ If $x = 16$, $\log_2x\cdot\log_2y = 4$ Of these, $4$ is the smallest. Thus the minimum is $4$. 

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