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Question: Find the maximum value of \(26 - {\left( {x + 7} \right)^2}\). A) \[24\] B) \[ - 23\] C) \[23\...

Find the maximum value of 26−(x+7)226 - {\left( {x + 7} \right)^2}.
A) 2424
B) −23 - 23
C) 2323
D) 2626

Explanation

Solution

Given problem tests, the concepts of derivatives and their applications. These type questions can be easily solved if we keep in mind the concepts of maxima and minima. In the problem, we are required to find the maximum value of the function in the variable xx, 26−(x+7)226 - {\left( {x + 7} \right)^2}. We do so by differentiating the function with respect to x and finding the critical points at which the function attains maximum or minimum values. Then, we use the second derivative test to check whether the critical point represents the minimum or maximum value of the function.

Complete step by step solution:
We have the function in x as 26−(x+7)226 - {\left( {x + 7} \right)^2}.
Let us assume the function 26−(x+7)226 - {\left( {x + 7} \right)^2} to be equal to y.
So, y=26−(x+7)2y = 26 - {\left( {x + 7} \right)^2}
Now, we differentiate both sides of the above equation. So, we get,
⇒dydx=ddx[26−(x+7)2]\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {26 - {{\left( {x + 7} \right)}^2}} \right]
We know the chain rule of differentiation as d[f(g(x))]dx=f′(g(x))×g′(x)\dfrac{{d\left[ {f\left( {g\left( x \right)} \right)} \right]}}{{dx}} = f'\left( {g\left( x \right)} \right) \times g'\left( x \right). Also, we know that derivative of any constant with respect to any variable is equal to zero. So, we get,
⇒dydx=−ddx[(x+7)2]\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{d}{{dx}}\left[ {{{\left( {x + 7} \right)}^2}} \right]
⇒dydx=−2(x+7)ddx[(x+7)]\Rightarrow \dfrac{{dy}}{{dx}} = - 2\left( {x + 7} \right)\dfrac{d}{{dx}}\left[ {\left( {x + 7} \right)} \right]
Now, we know that derivative of x with respect to x is 11. So, we get,
⇒dydx=−2(x+7)(1)\Rightarrow \dfrac{{dy}}{{dx}} = - 2\left( {x + 7} \right)\left( 1 \right)
⇒dydx=−2x−14−−−−(1)\Rightarrow \dfrac{{dy}}{{dx}} = - 2x - 14 - - - - \left( 1 \right)
Now, we substitute the value of y and equate the first derivative of the function to zero to find the critical points of the function. So, we get,
⇒−2x−14=0\Rightarrow - 2x - 14 = 0
Shifting the variable terms to right side of the equation, we get,
⇒−14=2x\Rightarrow - 14 = 2x
Dividing both sides by two, we get,
⇒x=−7\Rightarrow x = - 7
Now, we find the second derivative of the function. From equation (1)\left( 1 \right), we have,
⇒d2ydx2=dydx[−2x−14]\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{dy}}{{dx}}\left[ { - 2x - 14} \right]
Simplifying the expression, we get,
⇒d2ydx2=−2\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - 2
So, the second derivative of the function is negative for x=−7x = - 7. So, x=−7x = - 7 is the point of local maxima. So, the maximum value of function is attained at x=−7x = - 7.
So, y(−7)=26−(−7+7)2y\left( { - 7} \right) = 26 - {\left( { - 7 + 7} \right)^2}
⇒y(−7)=26−(0)2\Rightarrow y\left( { - 7} \right) = 26 - {\left( 0 \right)^2}
⇒y(−7)=26\Rightarrow y\left( { - 7} \right) = 26
Therefore, we get the maximum value of the function 26−(x+7)226 - {\left( {x + 7} \right)^2} as 2626. Hence, option (A) is the correct answer.

Note:
We can solve the problems involving the maxima and minima concept by two methods: the first derivative test and the second derivative test. The first derivative test helps in finding the local extremum points of the function. The second derivative test involves finding the local extremum points and then finding out which of them is local minima and which is local maxima.