Answer

Verified

87.9k+ views

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.

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.

Recently Updated Pages

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

Other Pages

Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main

Derive an expression for maximum speed of a car on class 11 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main