Question

If ${{x}^{2}}+{{y}^{2}}=25$ then the value of ${{\log }_{5}}\left[ Max(3x+4y) \right]$ is
A). 2
B). 3
C). 4
D). 5

Answer 1

Hint: Here, since the maximum function is given, first we have to find the maximum value of the function. For that we have to find the ${{1}^{st}}$ and ${{2}^{nd}}$ order derivatives of the function and substitute the first derivative to zero to find the maximum value and also the second derivative should be less than zero.

Complete step-by-step answer:
Here, we have the equation:
${{x}^{2}}+{{y}^{2}}=25$ …. (1)
Let $u=3x+4y$ …. (2)
Now, we have to find $y$ from equation (1). That is, by taking ${{x}^{2}}$ to the right hand side we get:
${{y}^{2}}=25-{{x}^{2}}$
By taking the square root on both the sides we get,
$y=\pm \sqrt{25-{{x}^{2}}}$
Since, we are considering the maximum value take
$y=\sqrt{25-{{x}^{2}}}$ ….. (3)
In the next step, let us substitute equation (3) in (2) we obtain:
$u=3x+4\sqrt{25-{{x}^{2}}}$
Next, by taking the derivative on both the sides with respect to x we get,
$\begin{aligned} & \dfrac{du}{dx}=\dfrac{d}{dx}\left( 3x+4\sqrt{25-{{x}^{2}}} \right) \\\ & \dfrac{du}{dx}=\dfrac{d}{dx}\left( 3x \right)+\dfrac{d}{dx}\left( 4\sqrt{25-{{x}^{2}}} \right) \\\ \end{aligned}$
$\dfrac{du}{dx}=\dfrac{d}{dx}(3x)+4\dfrac{d}{dx}\left( \sqrt{25-{{x}^{2}}} \right)$ …… (4)
We know that,
$\begin{aligned} & \dfrac{d}{dx}(3x)=3\dfrac{d}{dx}(x) \\\ & \dfrac{d}{dx}(3x)=3\times 1 \\\ \end{aligned}$
$\dfrac{d}{dx}(3x)=3\text{ }$ ….. (5)
Now, consider the derivative of $\sqrt{25-{{x}^{2}}}$, we will obtain:
$\begin{aligned} & \dfrac{d}{dx}\left( \sqrt{25-{{x}^{2}}} \right)=\dfrac{1}{2\sqrt{25-{{x}^{2}}}}\times \dfrac{d}{dx}\left( -{{x}^{2}} \right) \\\ & \dfrac{d}{dx}\left( \sqrt{25-{{x}^{2}}} \right)=\dfrac{1}{2\sqrt{25-{{x}^{2}}}}\times (-2x) \\\ \end{aligned}$
By cancelling 2 we get,
$\dfrac{d}{dx}\left( \sqrt{25-{{x}^{2}}} \right)=\dfrac{-x}{\sqrt{25-{{x}^{2}}}}$ ….. (6)
By substituting equation (5) and equation (6) in equation (4) we obtain:
$\dfrac{du}{dx}=3-\dfrac{4x}{\sqrt{25-{{x}^{2}}}}$
To find the maximum value, we have to substitute $\dfrac{du}{dx}=0$ i.e.
$\begin{aligned} & 3-\dfrac{4x}{\sqrt{25-{{x}^{2}}}}=0 \\\ & 3=\dfrac{4x}{\sqrt{25-{{x}^{2}}}} \\\ \end{aligned}$
By cross multiplication, the equation becomes:
$3\sqrt{25-{{x}^{2}}}=4x$
By squaring on both the sides we get,
$\begin{aligned} & {{\left[ 3\sqrt{25-{{x}^{2}}} \right]}^{2}}={{(4x)}^{2}} \\\ & 9(25-{{x}^{2}})=16{{x}^{2}} \\\ & 9\times 25+9\times -{{x}^{2}}=16{{x}^{2}} \\\ & 225-9{{x}^{2}}=16{{x}^{2}} \\\ \end{aligned}$
By taking $-9{{x}^{2}}$ on right side we get,
$\begin{aligned} & 225=16{{x}^{2}}+9{{x}^{2}} \\\ & 225=25{{x}^{2}} \\\ \end{aligned}$
By cancelling we obtain,
$9={{x}^{2}}$
Therefore, by taking the square root on both the sides we can say that,
$x=\pm 3$
For $x=\pm 3$, find the value of y from equation (1), i.e. equation (1) becomes:
$9+{{y}^{2}}=\text{ 25}$
Now, by taking 9 to the left side, 9 becomes -9. Hence, our equation becomes:
$\begin{aligned} & {{y}^{2}}=25-9 \\\ & {{y}^{2}}=16 \\\ \end{aligned}$
By taking square root on both the sides we get,
$y=\pm 4$
Next, we have to check that the second derivative of u is less than zero at $x=3$. For that find $\dfrac{{{d}^{2}}u}{d{{x}^{2}}}$.
$\begin{aligned} & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{du}{dx} \right) \\\ & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 3-\dfrac{4x}{\sqrt{25-{{x}^{2}}}} \right) \\\ & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{d}{dx}(3)-\dfrac{d}{dx}\left( \dfrac{4x}{\sqrt{25-{{x}^{2}}}} \right) \\\ & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=0-\dfrac{d}{dx}\left( \dfrac{4x}{\sqrt{25-{{x}^{2}}}} \right) \\\ \end{aligned}$
$\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{d}{dx}\left( \dfrac{4x}{\sqrt{25-{{x}^{2}}}} \right)$
Here, we have to apply the quotient rule. The quotient rule is:
${{\left( \dfrac{u}{v} \right)}^{\prime }}=\dfrac{v{u}'-u{v}'}{{{v}^{2}}}$
Hence, by applying quotient rule we will get:
$\begin{aligned} & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{\sqrt{25-{{x}^{2}}}\dfrac{d}{dx}(4x)-4x\dfrac{d}{dx}\left( \sqrt{25-{{x}^{2}}} \right)}{{{\left( \sqrt{25-{{x}^{2}}} \right)}^{2}}} \\\ & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{4\sqrt{25-{{x}^{2}}}\dfrac{d}{dx}(x)-4x\dfrac{-2x}{2\sqrt{25-{{x}^{2}}}}}{{{(25-x^{2})}}} \\\ & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{4\sqrt{25-{{x}^{2}}}+\dfrac{4{{x}^{2}}}{\sqrt{25-{{x}^{2}}}}}{{{(25-x^{2})}}} \\\ \end{aligned}$
By cross multiplication in the numerator of RHS we get
$\begin{aligned} & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{\dfrac{4\sqrt{25-{{x}^{2}}}\times \sqrt{25-{{x}^{2}}}+4{{x}^{2}}}{\sqrt{25-{{x}^{2}}}}}{25-{{x}^{2}}} \\\ & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{\dfrac{4(25-{{x}^{2}})+4{{x}^{2}}}{\sqrt{25-{{x}^{2}}}}}{25-{{x}^{2}}} \\\ & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{\dfrac{100-4{{x}^{2}}+4{{x}^{2}}}{\sqrt{25-{{x}^{2}}}}}{25-{{x}^{2}}} \\\ & \dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{\dfrac{100}{\sqrt{25-{{x}^{2}}}}}{25-{{x}^{2}}} \\\ \end{aligned}$
We know that $\dfrac{\dfrac{a}{b}}{c}=\dfrac{a}{bc}$.
Therefore, our equation becomes:
$\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{100}{\sqrt{25-{{x}^{2}}}\times 25-{{x}^{2}}}$
We also know that,
$\begin{aligned} & \sqrt{a+b)}(a+b)={{(a+b)}^{\dfrac{1}{2}}}(a+b) \\\ & \sqrt{a+b)}(a+b)={{(a+b)}^{\dfrac{1}{2}+1}} \\\ & \sqrt{a+b)}(a+b)={{(a+b)}^{\dfrac{3}{2}}} \\\ \end{aligned}$
By applying this in our equation we get,
$\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{100}{{{\left( 25-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}$
By putting $x=3$ in the above equation we get,
$\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=-\dfrac{100}{{{\left( 25-{{3}^{2}} \right)}^{\dfrac{3}{2}}}}<0$
Therefore, we can say that:
$\dfrac{{{d}^{2}}u}{d{{x}^{2}}}<0$ at $x=3$
Hence, the function $u=3x+4y$ is maximum at $x=3,y=3$. Therefore, we can write:
$\begin{aligned} & Max(3x+4y)=3\times 3+4\times 4 \\\ & Max(3x+4y)=9+16 \\\ & Max(3x+4y)=25 \\\ \end{aligned}$
Next, we have to find ${{\log }_{5}}\left[ Max(3x+4y) \right]$. i.e.
$\begin{aligned} & {{\log }_{5}}\left[ Max(3x+4y) \right]={{\log }_{5}}25 \\\ & {{\log }_{5}}\left[ Max(3x+4y) \right]={{\log }_{5}}{{5}^{2}} \\\ & {{\log }_{5}}\left[ Max(3x+4y) \right]=2{{\log }_{5}}5 \\\ & {{\log }_{a}}a=1 \\\ & \therefore {{\log }_{5}}\left[ Max(3x+4y) \right]=2 \\\ \end{aligned}$
Hence, the correct answer for this question is option (a)

Note: An alternate method to solve the problem is by substituting $x=5\cos \theta ,y=5\sin \theta$ in the equation $3x+4y$ .This method will reduce the steps, but you should be familiar with the trigonometric formulas.