用Mathematica求六元函数最值并显示自变量取值.FindMaxValue[{Sin[1/2 (Pi - x - y)] - ((u + x) Sin[(Pi - z - y)/2])/(z + q) - Sin[1/2 (Pi - u - p)] + (u Sin[1/2 (Pi - q - p)])/q,0 < x < \[Pi]/2,0 < y < \[Pi]/2,0 < z < \[Pi]/2,0 < u < \[Pi
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![用Mathematica求六元函数最值并显示自变量取值.FindMaxValue[{Sin[1/2 (Pi - x - y)] - ((u + x) Sin[(Pi - z - y)/2])/(z + q) - Sin[1/2 (Pi - u - p)] + (u Sin[1/2 (Pi - q - p)])/q,0 < x < \[Pi]/2,0 < y < \[Pi]/2,0 < z < \[Pi]/2,0 < u < \[Pi](/uploads/image/z/8632799-71-9.jpg?t=%E7%94%A8Mathematica%E6%B1%82%E5%85%AD%E5%85%83%E5%87%BD%E6%95%B0%E6%9C%80%E5%80%BC%E5%B9%B6%E6%98%BE%E7%A4%BA%E8%87%AA%E5%8F%98%E9%87%8F%E5%8F%96%E5%80%BC.FindMaxValue%5B%7BSin%5B1%2F2+%28Pi+-+x+-+y%29%5D+-+%28%28u+%2B+x%29+Sin%5B%28Pi+-+z+-+y%29%2F2%5D%29%2F%28z+%2B+q%29+-+Sin%5B1%2F2+%28Pi+-+u+-+p%29%5D+%2B+%28u+Sin%5B1%2F2+%28Pi+-+q+-+p%29%5D%29%2Fq%2C0+%3C+x+%3C+%5C%5BPi%5D%2F2%2C0+%3C+y+%3C+%5C%5BPi%5D%2F2%2C0+%3C+z+%3C+%5C%5BPi%5D%2F2%2C0+%3C+u+%3C+%5C%5BPi)
用Mathematica求六元函数最值并显示自变量取值.FindMaxValue[{Sin[1/2 (Pi - x - y)] - ((u + x) Sin[(Pi - z - y)/2])/(z + q) - Sin[1/2 (Pi - u - p)] + (u Sin[1/2 (Pi - q - p)])/q,0 < x < \[Pi]/2,0 < y < \[Pi]/2,0 < z < \[Pi]/2,0 < u < \[Pi
用Mathematica求六元函数最值并显示自变量取值.
FindMaxValue[{Sin[
1/2 (Pi - x - y)] - ((u + x) Sin[(Pi - z - y)/2])/(z + q) -
Sin[1/2 (Pi - u - p)] + (u Sin[1/2 (Pi - q - p)])/q,
0 < x < \[Pi]/2,0 < y < \[Pi]/2,0 < z < \[Pi]/2,0 < u < \[Pi]/2,
0 < p < \[Pi]/2,0 < q < \[Pi]/2},{x,y,z,u,p,q}]
用Mathematica求六元函数最值并显示自变量取值.FindMaxValue[{Sin[1/2 (Pi - x - y)] - ((u + x) Sin[(Pi - z - y)/2])/(z + q) - Sin[1/2 (Pi - u - p)] + (u Sin[1/2 (Pi - q - p)])/q,0 < x < \[Pi]/2,0 < y < \[Pi]/2,0 < z < \[Pi]/2,0 < u < \[Pi
要显示出位置的话可以使用FindMaximum或者NMaximize,但是你这个极值问题真正的麻烦在于……你这个式子在你所给的条件下根本就是趋于无穷的,你看看,在你的约束下,分母是可以为0的!
你的建模显然出了纰漏,先把你没分析进来的条件用上再说吧.