Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 12

The base costs $ab$, the glass front costs $2ac$, the wooden rear costs $ac$, and the other two walls cost $2bc$. Thus minimize $C=ab+3ac+2bc$ subject to $abc=V>0$. With the Lagrange multiplier convention $\nabla C=\lambda \nabla (abc)$, the equations areMultiply these by $a,b,c$, respectively. Comparing their equal right sides gives $ab=3ac=2bc$. Since all dimensions are positive, $a=2c$ and $b=3c$. The volume condition then yieldsThis stationary shape is the unique global minimum, not merely a candidate: the arithmetic-geometric mean inequality applied to the three cost terms giveswith equality exactly at those ratios. Therefore the least cost, in pounds in the specified units, isThis is a weighted open-box minimization, with each pair of opposite walls combined into its appropriate positive face weight.

The stationary point equations are $x^2+2y-21/4=0$ and $y^2+2x-21/4=0$. Subtracting them factors as $(x-y)(x+y-2)=0$, so all solutions lie either on $x=y$ or on $x+y=2$. The first case gives $x=y=3/2$ or $x=y=-7/2$; the second gives $(x,y)=(-1/2,5/2)$ or $(5/2,-1/2)$. This also verifies the two stated points and exhausts the possible solutions.

The Hessian matrix is $H=\left(\begin{array}{cc}2x& 2\\ 2& 2y\end{array}\right)$, with determinant $4(xy-1)$. The second-derivative test gives:

These extrema are local. Along $x=y=t$ the cubic term makes the function unbounded above and below, so it has no global maximum or minimum on the whole plane.