已知实数x,y满足3x^2+2y^2=6x,则x^2+y^2的最大值是?

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已知实数x,y满足3x^2+2y^2=6x,则x^2+y^2的最大值是?
已知实数x,y满足3x^2+2y^2=6x,则x^2+y^2的最大值是?

已知实数x,y满足3x^2+2y^2=6x,则x^2+y^2的最大值是?
解.3x²+2y²=6x
即3(x²-2x+1)+2y²=3
即(x-1)²+（2y²/3）=1
设x=1+cost,y=(√6/2)sint,则
x²+y²=1+2cost+cos²t+(3/2)sin²t
=1+2cost+cost²+(3/2)(1-cos²t)
=-(1/2)cos²t+2cost+5/2
=-(1/2)(cos²t-4cost+4)+9/2
=-(1/2)(cost-2)²+9/2
∵函数f(x)=-(1/2)(x-2)²+9/2是条开口向下,对称轴为x=2的抛物线,
∴f(x)在区间[-1,1]上单调递增
∵-1≤cost≤1
∴当cost=1时,x²+y²有最大值-(1/2)(1-2)²+9/2=4

给你个思路
将3x^2+2y^2=6x
化简成y^2=3x-3/2x^2
代入x^2+y^2中
再根据3x^2+2y^2=6x
x的取值范围求解最大值

3x^2+2y^2=6x
y^2=3x-3/2x^2
x^2+y^2=x^2+3x-3/2x^2=-1/2x^2+3x
-1/2*2x+3=0
x=3
-1/2*3^2+3*3=-9/2+9=4.5
x^2+y^2的最大值是4.5

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