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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 $$21 sin x + 72 cos x = \sqrt{(21)^2 + (72)^2}$$
= $$\sqrt{441 + 5184} = \sqrt{5625}$$
= $$75$$
