isi-entrance 2007 Q7

isi-entrance · India · solved Stationary points and optimisation Geometric or applied optimisation problem

Let $a$, $b$, $h$ be the three edges meeting at a particular vertex of a triangular prism, such that $a$, $b$ are sides of a base triangle with angle $\theta$ between them and $h$ is the height of the prism. Given that the total surface area is $K$, show that the volume $V$ satisfies $V \leq \sqrt{K^3/54}$, and find the dimensions of the prism of maximum volume.

Volume $V = \frac{1}{2}hab\sin\theta$. Surface area $K = \frac{1}{2}ab\sin\theta + ah + bh \geq 3\sqrt[3]{\frac{1}{2}ab\sin\theta \cdot ah \cdot bh} = \sqrt[3]{\frac{54(hab\sin\theta)^2}{\sin\theta}} \geq \sqrt[3]{54V^2}$. Hence $V \leq \sqrt{K^3/54}$. Equality holds when $\sin\theta = 1$ and $\frac{1}{2}ab\sin\theta = ah = bh$, i.e., $a = b = 2h$. Then $K = 6h^2 \Rightarrow h = \sqrt{K/6}$.

Let $a$, $b$, $h$ be the three edges meeting at a particular vertex of a triangular prism, such that $a$, $b$ are sides of a base triangle with angle $\theta$ between them and $h$ is the height of the prism. Given that the total surface area is $K$, show that the volume $V$ satisfies $V \leq \sqrt{K^3/54}$, and find the dimensions of the prism of maximum volume.

Paper Questions

Q1 Q3 Q5 Q7 Q8 Q10