Excuse me, how to solve a math application problem in primary school: there are 3 liters of water in a conical container, and the height of the water surface is exactly half that of the cone. How much can this container hold? Let the bottom area of a big cone be S, the height of the cone be H, the total volume be V, and the upper part be a small cone. The radius and height of a small cone are half that of a large cone, which is given by the cone volume formula V= 1/3. SH, the volume of the small cone V'= 1/3? × 1/4s× 1/2H = 1/8× 1/3？ SH= 1/8V, and the volume of a small cone is 1/8 of a large cone. So the volume of water is a big cone (1- 1/8)=7/8, so the total volume of the container =3÷(7/8)=24/7, so it can still hold water =24/7-3=3/7.