推 kane950544 :謝謝你 我也有用這個做法 但我搞不懂另一個作法的那 07/06 23:31
→ kane950544 :個步驟的邏輯 07/06 23:31
※ 引述《alasa15 (alasa)》之銘言：
: Find the extreme value of f(x,y)= x^4+y^4-4xy+1
: subject to the constraint x^2+y^2<=2
: ---------
f=x^4+y^4-4xy=(x^2+y^2)^2-2x^2y^2-4xy+1
=(x^2+y^2)^2-2(xy+1)^2+3
xy+1=r^2 sin2θ /2 +1
0<=r^2/2<=1
when r fixed
max f =r^4-2(1-r^2/2)^2+3
2*max = r^4+4r^2+2<=4+8+2
max=7
min f =r^4-2(1+r^2/2)^2+3
2*min = r^4-4r^2+2
=(r^2-2)^2-2>=-3
min =-1
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