I have a highly non linear profit function which depends on four independent variables (decision variables) E,W,T and p. I want to check concavity of profit function with respect to these four decision variables. To prove concavity, I have to prove that profit function's Hessian ( 4 *4 matrix) is negative (semi) definite for each values of decision variables but it is very difficult to draw any conclusion from its Hessian Matrix due to its complexity with many terms. Please guide whether any tool/technique/tricks or software package available for it. The profit function is given below ( it is written in Maple 2016.)-
Profit_function = (p*(W+1.00000000010^5W^.7*(.16-(.4-T)^2-0.3333333334e-4p^2(W^.3-E^.3))/p^2-E)+10E-20W-2.00000000010^6W^.7*(.16-(.4-T)^2-0.3333333334e-4p^2(W^.3-E^.3))/p^2-10-1.33333333310^5W^.7*(.4-((.4-T)^2+0.3333333334e-4p^2(W^.3-E^.3))^.5)^3/p^2+160000W^.7(.4-((.4-T)^2+0.3333333334e-4p^2(W^.3-E^.3))^.5)^2/p^2-(4*(W+1.00000000010^5W^.7*(.16-(.4-T)^2-0.3333333334e-4p^2(W^.3-E^.3))/p^2))(.4-((.4-T)^2+0.3333333334e-4p^2*(W^.3-E^.3))^.5)-(2*(W+E))(T-.4+((.4-T)^2+0.3333333334e-4p^2*(W^.3-E^.3))^.5)-5WT)/T
$\LaTeX$${\rm\LaTeX}$ translation
$(p*(W+10^5W^{0.7}(0.16-(0.4-T)^2-\frac 13 10^{-4}p^2(W^{0.3}-E^{0.3}))/p^2 \\ -E)+10E-20W-2\cdot 10^6W^{0.7}(0.16-(0.4-T)^2-\frac 13 10^{-4}p^2(W^{0.3} \\ -E^{0.3}))/p^2-10-\frac 43 10^5W^{0.7}(0.4-((0.4-T)^2+\frac 13 10^{-4}p^2(W^{0.3} \\ -E^{0.3}))^{0.5})^3/p^2+160000W^{0.7}(0.4-((0.4-T)^2+\frac 13 10^{-4}p^2(W^{0.3} \\ -E^{0.3}))^{0.5})^2/p^2-(4(W+10^5W^{0.7}(0.16-(0.4-T)^2 \\ -\frac 13 10^{-4}p^2(W^{0.3}-E^{0.3}))/p^2))(0.4-((0.4-T)^2 \\ +\frac 13 10^{-4}p^2(W^{0.3}-E^{0.3}))^{.5})-(2(W+E))(T-0.4+((0.4-T)^2 \\ +\frac 13 10^{-4}p^2(W^{0.3}-E^{0.3}))^{0.5})-5WT)/T$
Thanks and Regards, Nilesh
edit by YemonChoi: I have tried to tidy up the formatting; the previous version had line breaks in confusing places, and the grouping of various expressions seemed to obscure various patterns. I am leaving the OP's version above, in case I made any accidental errors in editing.
Inspecting this version, it looks like some changes of variables, such as $t:= 1-5T/2$, might be helpful; but I think it's best to leave that for a later edit, just in case my reformatting has introduced any errosr.
$$ \begin{aligned} T^{-1}\left[ p( W+10^5W^{0.7} p^{-2}(0.16-(0.4-T)^2-\frac 13 10^{-4}p^2(W^{0.3}-E^{0.3})) -E) \right. \\ +10E-20W \\ -2\cdot 10^6W^{0.7} p^{-2}\left(0.16-(0.4-T)^2-\frac 13 10^{-4}p^2(W^{0.3} -E^{0.3})\right) \\ -10 \\ -\frac{4}{3} 10^5W^{0.7} p^{-2}\left(0.4- \left[ (0.4-T)^2+\frac 13 10^{-4}p^2(W^{0.3} -E^{0.3}) \right]^{0.5}\right)^3 \\ +16\cdot 10^4 W^{0.7} p^{-2}\left(0.4-\left[ (0.4-T)^2 + \frac 13 10^{-4}p^2(W^{0.3}-E^{0.3})\right]^{0.5}\right)^2 \\ -4\left( W+10^5W^{0.7}p^{-2}\left[ 0.16-(0.4-T)^2 -\frac 13 10^{-4}p^2(W^{0.3}-E^{0.3}) \right] \right) \\ \times\left(0.4 - \left( (0.4-T)^2 +\frac 13 10^{-4}p^2(W^{0.3}-E^{0.3}) \right)^{0.5}\right) \\ -2(W+E) \left( T-0.4+\left( (0.4-T)^2 +\frac{1}{3} 10^{-4}p^2 (W^{0.3}-E^{0.3}) \right)^{0.5} \right) \\ \left. -5WT \right] \end{aligned} $$
