matlab求解二元函数最大值

y1=(p1-195)*3450.9*(2*(0.1645*832+129.05)/(0.1612*p1+121.61+0.1645*p2+129.05)).^1.6587*exp(-2*(0.1612*p1+121.61)/(0.1612*p1+121.61+0.1645*p2+129.05))/(exp(-2*(0.1612*p1+121.61)/(0.1612*p1+121.61+0.1645*p2+129.05))+exp(-2*(0.1645*p2+129.05)/(0.1612*p1+121.61+0.1645*p2+129.05)));

w1=(p2-372)*3450.9*(2*(0.1645*832+129.05)/(0.1612*p1+121.61+0.1645*p2+129.05)).^1.6587*exp(-2*(0.1645*p2+129.05)/(0.1612*p1+121.61+0.1645*p2+129.05))/(exp(-2*(0.1612*p1+121.61)/(0.1612*p1+121.61+0.1645*p2+129.05))+exp(-2*(0.1645*p2+129.05)/(0.1612*p1+121.61+0.1645*p2+129.05)));

求p1在（195，558），p2在（372，940）范围内，两个函数的最大值，以及相应的p1,p2值，我编出的程序直接显示

Warning: Trust-region-reflective algorithm does not solve this type of problem,
using active-set algorithm. You could also try the interior-point algorithm:
set the Algorithm option to 'interior-point' and rerun. 哪位大牛能不能帮忙编下程序啊，急求啊

我利用EXCEL的SOLVER算出来的值供你参考：
y1最大值=944317.200240146，当p1,p2取如下值时取得：
558.000000000
372.000000000

w1最大值=964143.177521792，当p1,p2取如下值时取得：
210.111434485846
940.00000000000