Question

Find the greatest value of the function $ y=7+2x\ln 25-{{5}^{x-1}}-{{5}^{2-x}} $

Answer 1

Hint: We will use the derivative method to find the greatest value of the function $ y=7+2x\ln x-{{5}^{x-1}}-{{5}^{2-x}} $ . We know that when we equate the first derivative to zero, we get the value of x at which function gives the minimum or maximum value. Now, if the double derivative at that point at which the first derivate is zero is less than zero then it will always give the maximum value.

Complete step by step answer:
We will use the concept of derivative to find the maximum value of the given function. We know that if any function gives the maximum value at ‘x’ then the first derivative at that point is equal to zero and the second derivative must be less than zero.
So, we will first find the first derivative of the given function.
Since, $ y=7+2x\ln 25-{{5}^{x-1}}-{{5}^{2-x}} $ , so upon differentiating both sides with respect to x we will get:
 $ \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( 7+2x\ln 25-{{5}^{x-1}}-{{5}^{2-x}} \right)}{dx} $
 $ \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( 7 \right)}{dx}+\dfrac{d\left( 2x\ln 25 \right)}{dx}-\dfrac{d\left( {{5}^{x-1}} \right)}{dx}-\dfrac{d\left( {{5}^{2-x}} \right)}{dx} $
Since, we know that $ \dfrac{d\left( {{a}^{x}} \right)}{dx}={{a}^{x}}\ln x $
 $ \Rightarrow \dfrac{dy}{dx}=2\ln 25-{{5}^{x-1}}\ln 5+{{5}^{2-x}}\ln 5 $
Now, we know that $ \ln 25=\ln {{5}^{2}}=2\ln 5 $ , so we will get:
 $ \Rightarrow \dfrac{dy}{dx}=2\times 2\ln 5-{{5}^{x-1}}\ln 5+{{5}^{2-x}}\ln 5 $
 $ \Rightarrow \dfrac{dy}{dx}=\ln 5\left( 4-{{5}^{x-1}}+{{5}^{2-x}} \right) $
Now, to get x for which function is maximum, we will equate $ \dfrac{dy}{dx}=0 $
 $ \Rightarrow \dfrac{dy}{dx}=\ln 5\left( 4-{{5}^{x-1}}+{{5}^{2-x}} \right)=0 $
 $ \Rightarrow \left( 4-{{5}^{x-1}}+{{5}^{2-x}} \right)=0 $

Now, we can write $ {{5}^{x-1}},{{5}^{2-x}} $ as $ \dfrac{{{5}^{x}}}{5},\dfrac{25}{{{5}^{x}}} $ .
 $ \Rightarrow \left( 4-\dfrac{{{5}^{x}}}{5}+\dfrac{25}{{{5}^{x}}} \right)=0 $
Now, let us assume that $ {{5}^{x}} $ as $ t $ .
So, we can write above equation as:
 $ \Rightarrow \left( 4-\dfrac{t}{5}+\dfrac{25}{t} \right)=0 $
 $ \Rightarrow \dfrac{20t-{{t}^{2}}+125}{5t}=0 $
 $ \Rightarrow \dfrac{{{t}^{2}}-20t-125}{5t}=0 $
Now, after splitting the middle term and then solving we will get:
 $ \Rightarrow \dfrac{{{t}^{2}}-25t+5t-125}{5t}=0 $
 $ \Rightarrow \dfrac{t\left( t-25 \right)+5\left( t-25 \right)}{5t}=0 $
 $ \Rightarrow \dfrac{\left( t+5 \right)\left( t-25 \right)}{5t}=0 $
 $ \therefore t=-5,25 $
Since, we have assumed $ {{5}^{x}} $ as $ t $ , so:
 $ {{5}^{x}}=-5,25 $
 $ \Rightarrow x=-1,2 $
Now, we will check at what point i.e. either at x = -1 or x = 2 we are getting maximum by using double derivative.
Since, $ \dfrac{dy}{dx}=\ln 5\left( 4-{{5}^{x-1}}+{{5}^{2-x}} \right)---\left( 1 \right) $
Now, we will differentiate (1) both sides with respect to ‘x’.
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( \ln 5\left( 4-{{5}^{x-1}}+{{5}^{2-x}} \right) \right)}{dx}\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\ln 5\left( \dfrac{d\left( 4 \right)}{dx}-\dfrac{d\left( {{5}^{x-1}} \right)}{dx}+\dfrac{d\left( {{5}^{2-x}} \right)}{dx} \right)\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\ln 5\left( -{{5}^{x-1}}\ln 5-{{5}^{2-x}}\ln 5 \right)\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{\left( \ln 5 \right)}^{2}}\left( {{5}^{x-1}}+{{5}^{2-x}} \right)--\left( 2 \right)\]
Since we know that the exponential function cannot give a negative value, and we have even power of $ \ln x $ so the double derivative is always negative.
Hence, at both the point i.e. at x = -1 and 2 we will have a local maximum so, we will solve for both the value of x and the x for which we get maximum will be the maximum value of the given function.
So, when x = -1;
 $ y=7+2\times \left( -1 \right)\ln 25-{{5}^{-1-1}}-{{5}^{2-\left( -1 \right)}} $
 $ \Rightarrow y=7-2\ln {{5}^{2}}-\dfrac{1}{{{5}^{2}}}-{{5}^{3}} $
 $ \Rightarrow y=7-2\times 2\ln 5-\dfrac{1}{25}-125 $
 $ \Rightarrow y=-118-4\ln 5-0.04 $
  $ \Rightarrow y=-118.04-4\ln 5 $
Now, when x = 2:
 $ y=7+2\times \left( 2 \right)\ln 25-{{5}^{2-1}}-{{5}^{2-2}} $
 $ \Rightarrow y=7+2\ln {{5}^{2}}-5-1 $
 $ \Rightarrow y=1+4\ln 5 $
We can see that when x = -1, $ y=-118.04-4\ln 5 $ which is smaller than the value that we get at x = 2 which is $ y=1+4\ln 5 $
So, we got the maximum value at x = 2 and the maximum value of the given function is $ 1+4\ln 5 $ . This is the required solution.

Note:
 When we get more than two-point for which the first derivative is zero, and the second is always positive or negative than to find maximum or minimum we are required to find the value of y for both all such x for which the first derivative is zero and then compare the result as per the requirement. If the second derivative is less than zero then we will get the minimum and if it is greater than zero then we will get the maximum.