Problem 411

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.