# 039 Review Problem - Bushels of wheat the grain elevator can hold

**Problem 39**

A grain elevator in the form of a frustum of a right circular cone is 24 ft. high, and the radii of its bases are 10 ft. and 5 ft., respectively; how many bushels of wheat will it hold if 1-1/4 cu. ft. equals 1 bu.?

**Solution 39**

$V = \frac{1}{3}\pi(R^2 + r^2 + Rr) \, h$

$V = \frac{1}{3}\pi[ \, 10^2 + 5^2 + 10(5) \, ](24)$

$V = 1400\pi ~ \text{ft.}^3$

$V = 1400\pi ~ \text{ft.}^3 \times \dfrac{1 ~ \text{bu.}}{1\frac{1}{4} ~ \text{ft.}^3}$

$V = 1120\pi ~ \text{bu.} = 3518.58 ~ \text{bu.}$ *answer*