Question

What is the maximum possible value of \((21 \sin x + 72 \cos x)\)?

• A. 21
• B. 57
• C. 63
• D. 75
• E. None of the above

Hint:

Always remember: \(\max(a \sin x + b \cos x) = \sqrt{a^2 + b^2}\). This shortcut saves time in trigonometric maximum-minimum problems.

Answer 1

The Correct Option is D

Step 1: Recall standard result.
The maximum value of an expression of the form: \[ a \sin x + b \cos x \] is given by: \[ \sqrt{a^2 + b^2} \]

Step 2: Apply values.
Here, \(a = 21\), \(b = 72\). So, \[ \text{Max Value} = \sqrt{21^2 + 72^2} \] \[ = \sqrt{441 + 5184} \] \[ = \sqrt{5625} = 75 \]

Step 3: Verification using derivative method.
Alternatively, let \[ f(x) = 21 \sin x + 72 \cos x \] Differentiate: \[ f'(x) = 21 \cos x - 72 \sin x \] Setting \(f'(x) = 0\): \[ 21 \cos x = 72 \sin x \quad \Rightarrow \quad \tan x = \frac{21}{72} = \frac{7}{24} \] So, \[ \sin x = \frac{7}{25}, \quad \cos x = \frac{24}{25} \] Substitute: \[ f(x) = 21 \cdot \frac{7}{25} + 72 \cdot \frac{24}{25} \] \[ = \frac{147}{25} + \frac{1728}{25} = \frac{1875}{25} = 75 \]

Final Answer:
\[ \boxed{75} \]