解斜三角形如图,铁路线AB段长100千
解: 设火车每吨·千米3a元,则公路每吨·千米5a元
总费用F=[3×BD+5×CD]×a
3×BD+5×CD=3(100-AD)+5CD
=300-3AD+5CD
CD=√(AC^+AD^)=√(400+AD^)
3×BD+5×CD=300-3AD+5√(400+AD^)=300+U
令-3AD+5√(400+AD^)=U
U+3AD=5√(400+AD^)
16AD^-6U×AD+10000-U^=0
∵AD≥0
∴△=36U^-4×16(1000-U^)≥0
U≥80
∴[U]min=80
[3×BD+5×CD]min=300+80=380
[F]min=380a
此时: U...全部
解: 设火车每吨·千米3a元,则公路每吨·千米5a元
总费用F=[3×BD+5×CD]×a
3×BD+5×CD=3(100-AD)+5CD
=300-3AD+5CD
CD=√(AC^+AD^)=√(400+AD^)
3×BD+5×CD=300-3AD+5√(400+AD^)=300+U
令-3AD+5√(400+AD^)=U
U+3AD=5√(400+AD^)
16AD^-6U×AD+10000-U^=0
∵AD≥0
∴△=36U^-4×16(1000-U^)≥0
U≥80
∴[U]min=80
[3×BD+5×CD]min=300+80=380
[F]min=380a
此时: U=80
16AD^-6U×AD+10000-U^=0
16AD^-480×AD+10000-8400=0
AD=15千米
。
收起