已知x,y为正实数,且满足关系式x^2-2x+4y^2=0,求xy的最大值。
答案是:
令xy=p
y=p/x
x^4-2x^3+4p^2=0
4p^2=2x^3-x^4=x^3(2-x)=27*(x/3)^3*(2-x)
耿昕回答:
x²-2x+4y²=0
→(x-1)²+y²/(1/2)²=1
故利用椭圆方程可设:x=1+cosθ,y=(1/2)sinθ
∴xy=(1/2)(1+cosθ)sinθ
=2sin(θ/2)cos³(θ/2)
→(xy)²=4·27·(1/3cos²(θ/2))·(1/3cos²(θ/2))·(1/3cos²(θ/2))·sin²(θ/2)
≤108·[(1/3cos²(θ/2)+1/3cos²(θ/2)+1/3cos²(θ/2)+sin²(θ/2))/4]^4
=108·[(sin²(θ/2)+cos²(θ/2))/4]^4
=27/64
∴所求最大值为:(xy)|max=3√3/8.
此时,sin²(θ/2)=1/3cos²(θ/2)→tan(θ/2)=√3/3,即θ=π/3.
代回所设得,x=3/2,y=√3/4.
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