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What will be maximum value of ${\text{3cos}}\theta {\text{ + 4sin}}\theta $
$
  \left( a \right){\text{ - 5}} \\
  \left( b \right){\text{ 5}} \\
  \left( c \right){\text{ 25}} \\
  \left( d \right){\text{ None of these}} \\
 $

Answer
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Hint-Use the concept of maxima and minima.Highest and lowest point is generally the maxima and minima of a graph.

Here we have to find the maximum value of ${\text{3cos}}\theta {\text{ + 4sin}}\theta $
So let ${\text{f}}\left( \theta \right) = {\text{3cos}}\theta {\text{ + 4sin}}\theta $
Now our first derivative ${{\text{f}}^1}\left( \theta \right) = - 3\sin \theta + 4\cos \theta $
Now double differentiating it we get ${{\text{f}}^{11}}\left( \theta \right) = - 3\cos \theta - 4\sin \theta $
In order to find max and min value we have to make ${{\text{f}}^1}\left( \theta \right) = 0$
Hence ${\text{ - 3sin}}\theta {\text{ + 4cos}}\theta {\text{ = 0}}$
On solving above we get ${\text{tan}}\theta {\text{ = }}\dfrac{4}{3}$
As we know that ${\text{tan}}\theta {\text{ = }}\dfrac{P}{H}$
Hence our ${\text{sin}}\theta {\text{ = }}\dfrac{4}{5}{\text{ and cos}}\theta {\text{ = }}\dfrac{3}{5}$
Now for this value of ${\text{sin}}\theta {\text{ and cos}}\theta $, the value of double derivative of $f(\theta ) $should be less than zero as we have to find maximum value of the expression.
${{\text{f}}^{11}}\left( \theta \right) < 0$
$ - 3\cos \theta - 4\sin \theta < 0$
Putting the values
$ - 3\left( {\dfrac{3}{5}} \right) - 4\left( {\dfrac{4}{5}} \right) < 0$
So max value of
${\text{3cos}}\theta {\text{ + 4sin}}\theta = 3 \times \left( {\dfrac{3}{5}} \right) + 4 \times \left( {\dfrac{4}{5}} \right)$
$ = \dfrac{{25}}{5}$Which is equal to 5
Hence option (b) is the right answer.
Note- Whenever we face such a problem the key concept that we need to use is that we always put the first derivative equal to 0 to obtain the values. Now double differentiate and cross verify that whether the value obtained corresponds to maximum or minimum for the function. This helps in reaching the right answer.