利用函數(shù)思想求體積的取值范圍（2022新高考Ⅰ，8）

2023-04-14 14:39 作者:數(shù)學(xué)老頑童 0人讀過 | 我要投稿

（2022新高考Ⅰ，8）已知正四棱錐的側(cè)棱長為 $l$ ，其各頂點(diǎn)都在同一球面上.若該球體的體積為 $36%5Cmathrm%7B%5Cpi%7D$ ，且 $3%5Cleqslant%20l%5Cleqslant%203%5Csqrt%7B3%7D$ ，則該四棱錐體積的取值范圍是（ ???）

A. $%5Cleft%5B%2018%2C%5Cfrac%7B81%7D%7B4%7D%20%5Cright%5D%20$

B. $%5Cleft%5B%20%5Cfrac%7B27%7D%7B4%7D%2C%5Cfrac%7B81%7D%7B4%7D%20%5Cright%5D%20$

C. $%5Cleft%5B%20%5Cfrac%7B27%7D%7B4%7D%2C%5Cfrac%7B64%7D%7B3%7D%20%5Cright%5D%20$

D. $%5Cleft%5B%2018%2C27%20%5Cright%5D%20$

解：設(shè)球 $O$ 的半徑為 $R$ ，

則 $%5Cfrac%7B4%7D%7B3%7D%5Cmathrm%7B%5Cpi%7DR%5E3%3D36%5Cmathrm%7B%5Cpi%7D$ ，解得 $%5Ccolor%7Bred%7D%7BR%3D3%7D$ .

如圖，設(shè)正四棱錐為 $P-ABCD$ ，其高為 $PH%3Dh$ .

設(shè)直線 $PH$ 與球面的另一交點(diǎn)為 $E$ ，連接 $AE$ 、 $HE$ .

易知 $%5Cbigtriangleup%20PHA%5Ctext%7B%E2%88%BD%7D%5Cbigtriangleup%20PAE$ ，

所以 $%5Cfrac%7BPH%7D%7BPA%7D%3D%5Cfrac%7BPA%7D%7BPE%7D$ ，即 $%5Cfrac%7Bh%7D%7Bl%7D%3D%5Cfrac%7Bl%7D%7B6%7D$ ，即 $%5Ccolor%7Bred%7D%7Bh%3D%5Cfrac%7Bl%5E2%7D%7B6%7D%7D$

所以

$%5Cbegin%7Baligned%7D%0A%09V_%7BP-ABCD%7D%26%3D%5Cfrac%7B1%7D%7B3%7D%5Ccdot%20h%5Ccdot%20S_%7B%5Ctext%7B%E6%AD%A3%E6%96%B9%E5%BD%A2%7DABCD%7D%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B3%7D%5Ccdot%20h%5Ccdot%20%5Cleft(%202%5Cleft%7C%20HA%20%5Cright%7C%20%5Cright)%20%5E2%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%0A%09%26%3D%5Cfrac%7B2%7D%7B3%7D%5Ccdot%20h%5Ccdot%20%5Cleft%7C%20HA%20%5Cright%7C%5E2%5C%5C%0A%09%26%3D%5Cfrac%7B2%7D%7B3%7D%5Ccdot%20h%5Ccdot%20%5Cleft(%20l%5E2-h%5E2%20%5Cright)%5C%5C%0A%09%26%3D%5Cfrac%7B2%7D%7B3%7D%5Ccdot%20%5Cfrac%7Bl%5E2%7D%7B6%7D%5Ccdot%20%5Cleft%5B%20l%5E2-%5Cleft(%20%5Cfrac%7Bl%5E2%7D%7B6%7D%20%5Cright)%20%5E2%20%5Cright%5D%5C%5C%0A%09%26%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7Bl%5E4%5Cleft(%2036-l%5E2%20%5Cright)%7D%7B324%7D%7D%5C%5C%0A%5Cend%7Baligned%7D$

令 $V%3Df%5Cleft(%20l%20%5Cright)%20%3D%5Cfrac%7Bl%5E4%5Cleft(%2036-l%5E2%20%5Cright)%7D%7B324%7D$ ，

其中 $l%5Cin%20%5Cleft%5B%203%2C3%5Csqrt%7B3%7D%20%5Cright%5D%20$ .則

$%5Cbegin%7Baligned%7D%0A%09f'%5Cleft(%20l%20%5Cright)%20%26%3D%5Cfrac%7B4l%5E3%5Ccdot%20%5Cleft(%2036-l%5E2%20%5Cright)%20%2Bl%5E4%5Ccdot%20%5Cleft(%20-2l%20%5Cright)%7D%7B324%7D%5C%5C%0A%09%26%3D%5Cfrac%7Bl%5E3%5Cleft(%202%5Csqrt%7B6%7D%2Bl%20%5Cright)%20%5Cleft(%20%5Ccolor%7Bred%7D%7B2%5Csqrt%7B6%7D-l%7D%20%5Cright)%7D%7B54%7D%5C%5C%0A%5Cend%7Baligned%7D$

令 $f%E2%80%99%5Cleft(%20l%20%5Cright)%20%3D0$ ，解得 $l%3D2%5Csqrt%7B6%7D$ ，

當(dāng) $l%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft%5B%203%2C2%5Csqrt%7B6%7D%20%5Cright)%20%7D$ ， $f'%5Cleft(%20l%20%5Cright)%20%3E0$ ， $f%5Cleft(%20l%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D$ ；

當(dāng) $l%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%202%5Csqrt%7B6%7D%2C3%5Csqrt%7B3%7D%20%5Cright%5D%20%7D$ ， $f%E2%80%99%5Cleft(%20l%20%5Cright)%20%3C0$ ， $f%5Cleft(%20l%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D$