Cho hình chóp tam giác S.ABC,SA=x,BC=y, các cạnh còn lại đều bằng 1.
1. Tính thể tích hình chóp theo x,y.
2. Với x,y nào thì thể tích hình chóp lớn nhất?
1. Gọi MN là trung điểm SABC.Do giả thiết, ΔABCCAS cân ở BC nên BMSA,CMSASA(BMC), hình chóp có thể tích
V=13SA.dtBMC=x3.dtBMC
Dễ thấy MB=MC=1x24 nên ΔBMC cânở MMNBC,
MN=MC2(y2)2=1x2+y24
dtMBC=12y1x2+y24 , V=xy61x2+y24
2. Ta có : x2+y242xy4=xy2
Vxy6.1xy2=16(xy)2.(2xy2)
V16.2.xy2.xy2.(2xy)162(xy2+xy2+2xy3)3
V16.1627=2327
V=2327{x2+y2=2xyxy/2=2xyx=y=23
Kết luận : Với x=y=23 thì V  lớn nhất

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