Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 20 / b

The triangular roof has height $(w/2)\tan \theta$ and two slopes of length $(w/2)\sec \theta$. Hence the area constraint and cost areThe wall height is fixed, so its cost is constant. Assume the physical case $A,h,\beta >0$. Minimize $C$ subject to the area constraint by Lagrange multipliers:Eliminating the positive common factors givesThus the optimum has $0<\theta <\pi /4$ andSet $D=A-wh>0$, so the area constraint gives $\tan \theta =4D/w^2$. Substituting into the preceding equation and multiplying by $w^3$ yields $w^4-16D^2=16hwD$. ThereforeFor a global check, the square of the total roof length isIts derivative has the sign of $P(w)=w^4+16Ahw-16A^2$. Since $P^{\prime }(w)>0$, $P(0)<0$ and $P(A/h)>0$, it has one root in this interval, with $Q$ decreasing before that root and increasing afterwards. This proves the unique global minimum, rather than merely a stationary design. The minimum roof length at fixed wall height and cross-sectional area is independent of $\alpha$ and of the positive factor $\beta$. If $\beta =0$, every feasible shape has equal cost; the displayed conditions then describe a possible minimizer but are not necessary.