Maximize using range

The function $f(x)=e^{\cos x} \left(380-x-x^2\right)=e^{\cos x} (19+x)(20-x)$ has two real roots, at $x=-19$ and $x=20$ .

Since $e^{\cos x}$ is always positive for real $x$ , the sign of $f$ is the same as the sign of $(19+x)(20-x)$ ; that is $f(x)>0$ precisely when $-19<x<20$ . The integral is therefore maximised by taking $a=-19$ , $b=20$ , giving $a+b=\boxed{-1}$ .

Let $f(a,b) = \int_{a}^{b} e^{\cos(x)} ( 380-x-x^2) dx$ , and let us compute $grad f = 0.$ This yields:

$\frac{\partial{f}}{\partial{a}} = -e^{\cos(a)} (380-a-a^2) = -e^{\cos(a)} (19-a)(20+a) = 0 \Rightarrow a = -20, 19;$

$\frac{\partial{f}}{\partial{b}} = e^{\cos(b)} (380-b-b^2) = e^{\cos(a)} (19-a)(20+a) = 0 \Rightarrow b = -20, 19;$

Since we require $a \le b$ we have three possible critical points to check: $A(-20,-20); B(19,19); C(-20,19)$ . Clearly $f$ equals zero at both $A$ and $B,$ which leaves us $C.$ If we compute the Hessian matrix of $f$ at $C$ :

$F(a,b) = \begin{pmatrix} f_{aa} & f_{ab} \\ f_{ba} & f_{bb} \end{pmatrix} = \begin{pmatrix} e^{\cos(a)} ((380-a-a^2)\sin(a) + 1 + 2a)& 0 \\ 0 & e^{\cos(b)} (-(380-b-b^2)\sin(b) -1 - 2b) \end{pmatrix}$ ;

or $F(-20,19) = \begin{pmatrix} -39e^{\cos(-20)} & 0 \\ 0 & -39e^{\cos(19)} \end{pmatrix}$

which is negative-definite since the Hessian contains two real, distinct negative eigenvalues. Hence, point $C$ maximizes $f \Rightarrow a + b = 19 - 20 = \boxed{-1}.$