XAT 2011 Question 72

Question 72

What is the maximum possible value of (21 Sin X + 72 Cos X)?

Solution

To find the minimum or maximum value, we need to find first derivative and put it equal to zero

Expression : $$f(x) = 21 sin x + 72 cos x$$

=> $$f'(x) = 21 cos x - 72 sin x = 0$$

=> $$21 cos x = 72 sin x$$

=> $$\frac{sin x}{cos x} = tan x = \frac{21}{72} = \frac{7}{24}$$

=> $$sin x = \frac{7}{\sqrt{7^2 + 24^2}} = \frac{7}{\sqrt{49 + 576}}$$

=> $$sin x = \frac{7}{\sqrt{625}} = \frac{7}{25}$$

Similarly, $$ cos x = \frac{24}{25}$$

Putting it in the original expression, we get :

=> $$f(x) = (21 \times \frac{7}{25}) + (72 \times \frac{24}{25})$$

= $$\frac{147}{25} + \frac{1728}{25} = \frac{1728 + 147}{25}$$

= $$\frac{1875}{25} = 75$$


Shortcut Method : Maximum value of $$a sin x + b cos x = \sqrt{a^2 + b^2}$$

and minimum value = $$- \sqrt{a^2 + b^2}$$

=> Max value of $$21 sin x + 72 cos x = \sqrt{(21)^2 + (72)^2}$$

= $$\sqrt{441 + 5184} = \sqrt{5625}$$

= $$75$$



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