由于某种产品的原料提价，因此厂家决定对产品进行提价，现有三种方案.方案1：第一次提价p%，第二次提价q%；方案2：第一次提价q%，第二次提价p%；方案3：两次均提价$\dfrac {p+q} {2}\% $；其中p，q是不相等的正数.问：三种方案哪种提价最多？

不妨把产品原价看作$1$，设$p\% =x$,$q\% =y$.

方案$1$提价为$\left ( {1+p\% } \right )\left ( {1+q\% } \right )=\left ( {1+x} \right )\left ( {1+y} \right )$;

方案$2$提价为$\left ( {1+q\% } \right )\left ( {1+p\% } \right )=\left ( {1+y} \right )\left ( {1+x} \right )$;

方案$3$提价为$\left ( {1+\dfrac {p+q} {2}\% } \right )\left ( {1+\dfrac {p+q} {2}\% } \right )=\left ( {1+\dfrac {x+y} {2}} \right )\left ( {1+\dfrac {x+y} {2}} \right )$.

显然$\left ( {1+p\% } \right )\left ( {1+q\% } \right )=\left ( {1+q\% } \right )\left ( {1+p\% } \right )$.

$\because \left ( {1+\dfrac {x+y} {2}} \right )\left ( {1+\dfrac {x+y} {2}} \right )-\left ( {1+x} \right )\left ( {1+y} \right )$

$=1+\dfrac {x+y} {2}+\dfrac {x+y} {2}+\left ( {\dfrac {x+y} {2}} \right )^{2}-1-x-y-xy$

$=1+x+y+\left ( {\dfrac {x+y} {2}} \right )^{2}-1-x-y-xy$

$=\left ( {\dfrac {x+y} {2}} \right )^{2}-xy$

$=\left ( {\dfrac {x-y} {2}} \right )^{2}\gt 0$,

$\therefore $方案$3$提价最多.