Problem 568

The water is in the shape of a rectangular prism with a 2 cm by 5
cm base and depth 6 cm.

Therefore, the volume of water is $2 \text{ cm} \times 5 \text{ cm} \times 6\text{ cm} = 60\text{ cm}^3$.

A face of the prism having the greatest area has dimensions 5 cm by 8
cm.

When the prism is tipped so that it stands on a 5 cm by 8 cm face, the
water is once again in the shape of a rectangular prism with a 5 cm by 8
cm base and unknown depth.

Suppose that after the prism is tipped, the water’s depth is $d$ cm.

Since the volume of water is still $60\text{ cm}^3$ when the prism is tipped, then $5 \text{ cm} \times 8 \text{ cm} \times d\text{ cm} = 60\text{ cm}^3$ or $40d\text{ cm}^3 = 60\text{ cm}^3$, and so $d=\dfrac{60}{40}=\dfrac32$.

When the prism is tipped so that it stands on a face with the greatest
area, the depth of the water is $\dfrac32 \text{ cm}=1.5$ cm.