A barn is to be constructed with a uniform cross-sectional area A throughout its length. The cross-section is to be a rectangle of wall height h (fixed) and width w, surmounted by an isosceles triangular roof that makes an angle θ with the horizontal. The cost of construction is α per unit height

Question

A barn is to be constructed with a uniform cross-sectional area A throughout its length. The cross-section is to be a rectangle of wall height h (fixed) and width w, surmounted by an isosceles triangular roof that makes an angle θ with the horizontal. The cost of construction is α per unit height of wall and β per unit (slope) length of roof. Show that, irrespective of the values of α and β, to minimise costs w should be chosen to satisfy the equation

$w^4$ = 16A(A − wh),

and θ made such that 2 tan 2θ = w/h.

[latex]w^4[/latex] = 16A(A − wh),

and θ made such that 2 tan 2θ = w/h.

Accepted Answer

The cost always includes 2αh for the vertical walls, which can therefore be ignored in the minimisation procedure. The rest of the calculation will be solely concerned with minimising the roof area, and the optimum choices for w and θ will be independent of β, the actual cost per unit length of the roof.

The cost of the roof is [latex]2 \beta imes \frac{1}{2} w \sec heta[/latex] , but w and θ are constrained by the requirement that

[latex]A=w h+\frac{1}{2} w \frac{w}{2} an heta.[/latex]

So we consider G(w, θ), where

[latex]G(w, heta)=\beta w \sec heta-\lambda\left(w h+\frac{1}{4} w^2 an heta ight),[/latex]

and the implications of equating its partial derivatives to zero. The first derivative to be set to zero is

[latex]\begin{aligned}\quad \frac{\partial G}{\partial heta} & =\beta w \sec heta an heta-\frac{\lambda}{4} w^2 \sec ^2 heta, \\Rightarrow \quad 0 & =\beta \sin heta-\frac{1}{4} \lambda w, \\Rightarrow \quad \lambda & =\frac{4 \beta \sin heta}{w} .\end{aligned}[/latex]

A second equation is provided by differentiation with respect to w and yields

[latex]\frac{\partial G}{\partial w}=\beta \sec heta-\lambda h-\frac{1}{2} \lambda w an heta.[/latex]

Setting ∂G/∂w = 0, multiplying through by cos θ and substituting for λ, we obtain

[latex]\begin{aligned}\beta-2 \beta \sin ^2 heta & =\frac{4 \beta \sin heta h \cos heta}{w}, \w \cos 2 heta & =2 h \sin 2 heta, \ an 2 heta & =\frac{w}{2 h} .\end{aligned}[/latex]

This is the second result quoted.

The overall area constraint can be written

[latex] an heta=\frac{4(A-w h)}{w^2}.[/latex]

From these two results and the double angle formula [latex] an 2 \phi = 2 an \phi/(1− an^2 \phi)[/latex], it follows that

[latex]\begin{aligned}\frac{w}{2 h} & = an 2 heta \& =\frac{\frac{8(A-w h)}{w^2}}{1-\frac{16(A-w h)^2}{w^4}}, \16 w h(A-w h) & =w^4-16(A-w h)^2, \w^4 & =16 A(A-w h) .\end{aligned}[/latex]

This is the first quoted result, and we note that, as expected, both optimum values are independent of β.

Question 5.19: A barn is to be constructed with a uniform cross-sectional a...

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