In the article, we prove that λ1=1/2+[(2+log(1+2))/2]1/ν−1/2 $\lambda _{1}=1/2+\sqrt{ [ (\sqrt{2}+ \log (1+\sqrt{2}) )/2 ]^{1/\nu }-1}/2$, μ1=1/2+6ν/(12ν) $\mu _{1}=1/2+\sqrt{6 \nu }/(12\nu )$, λ2=1/2+[(π+2)/4]1/ν−1/2 $\lambda _{2}=1/2+\sqrt{ [(\pi +2)/4 ] ^{1/\nu }-1}/2$ and μ2=1/2+3ν/(6ν) $\mu _{2}=1/2+\sqrt{3\nu }/(6\nu )$ are the best possible parameters on the interval [1/2,1] such that the double inequalities Cν[λ1x+(1−λ1)y,λ1y+(1−λ1)x]A1−ν(x,y)<RQA(x,y)<Cν[μ1x+(1−μ1)y,μ1y+(1−μ1)x]A1−ν(x,y),Cν[λ2x+(1−λ2)y,λ2y+(1−λ2)x]A1−ν(x,y)<RAQ(x,y)<Cν[μ2x+(1−μ2)y,μ2y+(1−μ2)x]A1−ν(x,y) $$\begin{aligned}& C^{\nu }\bigl[\lambda _{1}x+(1-\lambda _{1})y, \lambda _{1}y+(1-\lambda _{1})x\bigr]A ^{1-\nu }(x, y) \\& \quad < \mathcal{R}_{QA}(x, y)< C^{\nu }\bigl[\mu _{1}x+(1-\mu _{1})y, \mu _{1}y+(1-\mu _{1})x\bigr]A^{1-\nu }(x, y), \\& C^{\nu }\bigl[\lambda _{2}x+(1-\lambda _{2})y, \lambda _{2}y+(1-\lambda _{2})x\bigr]A ^{1-\nu }(x, y) \\& \quad < \mathcal{R}_{AQ}(x, y)< C^{\nu }\bigl[\mu _{2}x+(1-\mu _{2})y, \mu _{2}y+(1-\mu _{2})x\bigr]A^{1-\nu }(x, y) \end{aligned}$$ hold for all x,y>0 with x≠y and ν∈[1/2,∞) $\nu \in [1/2, \infty )$, where A(x,y) $A(x, y)$ is the arithmetic mean, C(x,y) $C(x, y)$ is the contraharmonic mean, and RQA(x,y) $\mathcal{R}_{QA}(x, y)$ and RAQ(x,y) $\mathcal{R}_{AQ}(x, y)$ are two Neuman means.