# What is the maximum value of y = 3*cosx + 4*sinx + 10. Shouldn't the answer be a simple 14.

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### 2 Answers

You are required to find the maximum value of `y = 3*cosx + 4*sinx + 10` .

One way of doing this is with calculus. For a function y = f(x), the maximum value lies at the solution x = a of f'(x) = 0 and if f''(a) is negative.

For `y = 3*cosx + 4*sinx + 10`

`y' = 3*(-sin x) + 4*cos x`

`3*(-sin x) + 4*cos x = 0`

=> `4*cos x = 3*sin x`

=> `tan x = 4/3`

`x = tan^-1(4/3)`

`y'' = -3*cos x - 4*sin x`

At `x = tan^-1(4/3)` , y'' is negative

Substituting `x = tan^-1(4/3)` in `y = 3*cosx + 4*sinx + 10` gives `3*cos (tan^-1(4/3)) + 4*sin (tan^-1(4/3)) + 10 = 15`

**The maximum value of `y = 3*cosx + 4*sinx + 10` is 15.**

P.S. The answer to the question is not 14 as the greatest ratio of sin x to cos x is not 1. There are values of x for which this ratio is greater than 1.

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