1 設(shè)對(duì)角線長(zhǎng)為2a,夾角為θ,則知一邊長(zhǎng)為
2a*sin(θ/2)另一邊長(zhǎng)為2a*cos(θ/2)
則
矩形面積S=2a*sin(θ/2)*2a*cos(θ/2)=2a^2*[2sin(θ/2)cos(θ/2)]=2a^2*sinθ
又0≤sinθ≤1,當(dāng)且僅當(dāng)sinθ=1,θ=90°時(shí)
有Smax=2a^2
此時(shí)矩行為正方形
矩形周長(zhǎng)C=2a*sin(θ/2)+2a*cos(θ/2)
=2a[sin(θ/2)+cos(θ/2)]
=2√2*asin(θ/2+45°)
又0≤sin(θ/2+45°)≤1,
當(dāng)且僅當(dāng)sin(θ/2+45°),θ/2+45°=90°,θ=90°時(shí)
有Cmax=2√2*a
此時(shí)矩形為正方形
2 顯然滿足條件的圓柱被經(jīng)過(guò)圓心且平行于底面的平面平分為兩部分
則圓柱底面積=πr²
h=2√(R²-r²)
V=πr²*2√(R²-r²)=4π√[(r²/2)²(R²-r²)]
根據(jù)均值不等式
(R²/3)³=[(r²/2+r²/2+R²-r²)/3]³≥(r²/2)²(R²-r²)
當(dāng)r²/2=R²-r²時(shí)取等號(hào)
此時(shí)r=√6R/3,h=2√3R/3