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June 7th, 2015, 06:43 PM   #1
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find minimum with log

Hi, I need help with this two problems. Please someone show me the way to solve them! Thanx
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June 7th, 2015, 08:37 PM   #2
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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.
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June 7th, 2015, 10:09 PM   #3
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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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