[The volume of a square-based pyramid is \(\frac{1}{3} \times \text{base area} \times \text{height}\

Question

[The volume of a square-based pyramid is $\frac{1}{3} \times \text{base area} \times \text{height}$ .]

A manufacturer makes open containers in the shape of an inverted square pyramid with square base of side length $l$ cm, vertical height $h$ cm and fixed slant edge $k$ cm, as shown above. He decides that the volume of each container, $V\text{ cm}^3$, should be as large as possible.

Show that $V^2 = \frac{k^2l^4}{9} - \frac{l^6}{18}$.[2]
Use differentiation to find, in terms of $k$, the exact maximum value of $V$, proving that it is a maximum.[6]
Water is poured into an empty container with the dimensions found in part (b) at a constant rate of $\frac{1}{3}\text{ cm}^3$ per second. It is given that at time $t$ seconds, the volume of water in the container is $W\text{ cm}^3$ and the water level is $p$ cm. Show that $W = \frac{4}{3}p^3$ and hence, find the rate of increase of $p$ at the instant when $p = \frac{k}{4}$.[4]

Accepted Answer

(a) $V^2=\dfrac{k^2l^4}{9}-\dfrac{l^6}{18}$ (b) $V_{\max}=\dfrac{4k^3}{9\sqrt3}$ (c) $W=\dfrac43p^3$, $\dfrac{\mathrm dp}{\mathrm dt}=\dfrac4{3k^2}\text{ cm s}^{-1}$

2024 VJC Promo Q12

2024 VJC Promo Q12

Video Solution:

Solution:

Solution locked