Maximizing a Logarithm

Since $\ln(a)$ is a monotonically increasing function, to maximize $\ln(a)$ we must maximize $a$ . Thus, we must maximize $y^2+4xy+7$ .

We can write $x+y=\dfrac{1}{4}\implies 4x=1-4y$ Substituting, $y^2+4xy+7=-3y^2+y+7$ However, we know how to maximize a quadratic. The maximum occurs when $y=\dfrac{-1}{2(-3)}=\dfrac{1}{6}$ , and since $x+y=\dfrac{1}{4}$ , $x=\dfrac{1}{12}$ . Finally, $xy=\dfrac{1}{6\cdot 12}=\dfrac{1}{72}$ , and the answer is $\boxed{72}$ .

We rearrange the equation $x+y=\frac{1}{4}$ to get $x=-y+\frac{1}{4}.$ Now we substitute this into the logarithm expression. $\begin{aligned}\ln(y^2+4xy+7)&=\ln\left(y^2+4y\left(-y+\frac{1}{4}\right)+7\right)\\&=\ln(y^2-4y^2+y+7)\\&=\ln(-3y^2+y+7)\end{aligned}.$ Since the natural log is an increasing function, $\ln(a)\geq\ln(b)\text{ for }a\geq b.$ Thus, our logarithmic expression is maximized when the expression inside is maximized, which is a quadratic. The vertex of the downward opening parabola is $\begin{aligned}y&=-\frac{b}{2a}=\frac{1}{6}\\ \implies x&=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\end{aligned}.$ Finally, our desired answer is $\frac{1}{xy}=\frac{1}{\frac{1}{6}\cdot\frac{1}{12}}=\boxed{72}.$

Note that since $\ln(x)$ is increasing for all $x$ , the expression attains its maximum when $y^2+4xy+7$ attains its maximum. Note that $y^2+4xy=\frac{1}{3}(3y)(y+4x)\le \frac{1}{12}((3y)+(y+4x))^2=\frac{1}{12}$ since $ab\le \frac{1}{4}(a+b)^2$ by AM-GM. Thus $y^2+4xy+7$ attains its maximum when $3y=y+4x$ , which is when $y=2x$ , so $y=\frac{1}{6}$ and $x=\frac{1}{12}$ . It follows immediately that the answer is $6*12=\boxed{72}$

$In( y^2 + 4xy + 7 ) = log_e( y^2 + 4xy + 7 )$

Since $e$ > 1 , $y^2 + 4xy + 7$ will have to be the maximal for $In( y^2 + 4xy + 7 )$ to be the maximal.

Since $x = \frac {1}{4} - y$ , we substitute x for y, we can get $y^2 + 4xy + 7 = -3 y^2 + y + 7.$

The maximum point for the expression is $-\frac {b}{2a} = - \frac {1}{-6} = \frac {1}{6}$

When y = $\frac {1}{6}, x = \frac {1}{4} - y = \frac {1}{12}$ .

The value of $\frac {1}{xy}$ will be $\frac {1}{\frac {1}{6} \cdot \frac {1}{12}} = 72$

From the given equation, we have, $x=\frac{1}{4}-y$ . Substituting this into $y^2+4xy+7$ , we get, $y^2+4y(\frac{1}{4}-y)+7 =y-3y^2+7 =3y(\frac{1}{3}-y)+7$ .

Here, $\ln(y^2+4xy+7)=\ln(3y(\frac{1}{3}-y)+7)$ will be maximum when $3y(\frac{1}{3}-y)+7$ will be maximum. Again, this will be maximum when $y(\frac{1}{3}-y)$ will be maximum.

Now, from the AM-GM Inequality, we have, $\sqrt{y(\frac{1}{3}-y)}\leq \frac{y+\frac{1}{3}-y}{2}\Rightarrow y(\frac{1}{3}-y)\leq \frac{1}{36}$ .

So, the maximum value of $y(\frac{1}{3}-y)$ will be $\frac{1}{36}$ , and, this value will be achieved only when equality is established. There will be equality if and only if, $y=\frac{1}{3}-y$ , i.e, $y=\frac{1}{6}$ . So, the given expression will attain its maximum when $y=\frac{1}{6}$ . And, then, $x=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$ .

Therefore, when the given expression is maximum, $xy=\frac{1}{6}.\frac{1}{12}=\frac{1}{72}\Rightarrow\frac{1}{xy}=\boxed{72}$

To maximize the logarithm, we need to maximize the function inside the logarithm, because the value of a logarithm increases as the value of its operand increases.

$x + y = \dfrac{1}{4}$

$x = \dfrac{1}{4} - y$

Substituting this expression for $x$ in the quadratic:

$y^{2} + 4\left(\dfrac{1}{4} - y\right)y + 7$

$\Rightarrow y^{2} + y - 4y^{2} + 7$

$\Rightarrow -3y^{2} + y + 7$

The graph of this quadratic is an upside-down parabola, due to the negative coefficient on $y^{2}$ . Hence, the quadratic has a maximum value, not a minimum one. This maximum value occurs at:

$\dfrac{d}{dy}\left(-3y^{2} + y + 7\right) = 0$

$\Rightarrow -6y + 1 = 0$

$\Rightarrow y = \dfrac{1}{6}$

Then,

$x = \dfrac{1}{4} - \dfrac{1}{6} = \dfrac{1}{12}$

$\Rightarrow xy = \dfrac{1}{6} \times \dfrac{1}{12} = \dfrac{1}{72}$

$\Rightarrow \dfrac{1}{xy} = \boxed{72}$

given x + y = 4. from this x = 1/4 - y

substituting this in ln(y^2 + 4y +7), we get ln (-3y^2 + y + 7)

for ln (any polynomial to be maximal,we should differentiate the given polynomial and equate it to zero),

on differentiating -3y^2 = y =7, we get, -6y + 1 = 0 so, y = 1 /6 so ,x = 12

so,1/xy = 1/(1/6 * 1/12) so, 1/xy = 72

The function ln(...) restricted to the line x+y = 1/4 admits a maximum at x = 1/12; hence y = 1/4 - 1/12.

x+y = 1/4 => x = 1/4 - y; substituting x in y^2 + 4xy + 7 and finding the value of y to maximise the function; we get, y=1/6. Therefore, x=1/12 and xy=72

differentiate ln(y^{2} +4xy+7) and x+y=1/4
we get, {1/(y^{2}+4xy+7)}*(2yy'+4xy'+4y+0) eq(3) and, 1+y'=0 y'=-1 now put value of y' in eq(3) and find the result

Substitute x = ( 1/4 - y ) in the expression ln (y^2 + 4xy + 7) The expression becomes ln (7 + y - 3 y^2). For this expression to be maximum, the differential wrt y should be 0. This gives y = 1/6, and x = 1/12. Therefore, 1/xy = 72

We are given $x+y=\frac{1}{4}$ and $f(y)= ln(y^{2} + 4xy +7)$

$=> x=\frac{1-4y}{4}$

Substituting x into $f(y)= ln(y^{2} + 4xy +7)$

$=> f(y) = ln(y^{2} + (1-4y)y + 7)$

$=> f(y) = ln(-3^{2} + y + 7)$

$=> \frac{d(f(y)}{dy} = \frac{1}{-3y^{2}+y+7}$

Putting the value of $\frac{d(f(y)}{dy}$ = 0

we get,

$y=\frac{1}{6}$

and, $x=\frac{1}{12}$

Therefore, $\frac{1}{xy} = \boxed{72}$

With differential Method we've y^2 + 4y(1/4 -y) + 7 become -6y + 1 = 0 "to find maximum value). So, -6y = -1 --> y =1/6.Then x = 1/4 - 1/6 = 1/12. Thus, 1/x.y = 1/(1/12.1/6) = 1/(1/72) = 72. answer : 72