$解：?(1)M(12，??0)，??P(6，??6)?$
$? (2)?設(shè)二次函數(shù)表達式為?y=a(x-6)^2+6?$
$∵函數(shù)?y=a(x-6)^2+6?的圖像經(jīng)過點?(0，??0)?$
$∴?0=a(0-6)^2+6，?即?a=- \frac {1}{6}?$
$∴拋物線相應(yīng)的函數(shù)表達式為?y=-\frac {1}{6} (x-6)^2+6，?$
$即?y=-\frac 16x^2+2x ?$
$?(3)?設(shè)?A(m，??0)，?則?B(12-m，??0)，??C(12-m，??- \frac {1}{6}\ \mathrm {m^2}+2\ \mathrm {m})，$
$??D(m，??- \frac {1}{6}\ \mathrm {m^2} +2\ \mathrm {m})?$
$∴“支撐架”總長?AD+DC+CB= (- \frac {1}{6}\ \mathrm {m^2}+2m)+(12-2m) +(- \frac {1}{6}\ \mathrm {m^2}+2m )?$
$? =- \frac {1}{3}\ \mathrm {m^2}+2m+12=- \frac {1}{3} (m-3)^2+15?$
∵此二次函數(shù)的圖像開口向下
$∴當(dāng)?m=3\ \mathrm {m} ?時，?AD+DC+CB?有最大值為?15\ \mathrm {m}?$
