Fencing a Rectangular Pen Along a River
A farmer has 120 m of fencing to build a rectangular pen along a straight river. The river forms one side of the pen, so no fence is needed there. A tractor must be able to enter, so the width perpendicular to the river must be at least 10 m. The property line limits the side parallel to the river to at most 45 m, and the farmer wants that side to be at least 20 m to keep the gate layout practical. The farmer can use all or part of the fencing, but leftover fencing has no value for this project. A neighbour claims the best shape is always the one where the width is half the length, because that maximizes area for a fixed perimeter. The farmer wants a recommendation that respects the constraints and a check of the neighbour’s claim under these conditions.
1. Let x be the width perpendicular to the river and y the length parallel to it. Write the fencing equation and the area function. State the feasible domain for x given all constraints.
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Let x be width perpendicular to the river and y the length parallel to it. Fencing is needed on two widths and one length, so 2x+y=120, giving y=120-2x. The area is A(x)=xy=x(120-2x)=120x-2x^2. Constraints: x≥10; 20≤y≤45. Substituting y=120-2x gives 20≤120-2x≤45, so 37.5≤x≤50. Thus the feasible domain is 37.5≤x≤50 (the x≥10 condition is automatically satisfied).
2. Find the dimensions that maximize the area under the constraints. Compare with the neighbour’s unconstrained 'width half length' claim and explain any difference.
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The unconstrained vertex of A(x)=120x-2x^2 is x=-120/(2×-2)=30, giving y=120-60=60 and area 1800 m². This violates y≤45. On the feasible domain [37.5,50], the parabola opens downward with vertex at x=30, which is to the left of the domain, so A is decreasing throughout the domain. Therefore the maximum occurs at the smallest allowed x, x=37.5, with y=45. Area=37.5×45=1687.5 m². The neighbour's 'width half length' shape (x=30,y=60) is not feasible and is not the constrained optimum. The property-line limit forces a wider, shorter pen.
3. The farmer considers dividing the same total rectangular area into two identical adjacent pens with a shared fence perpendicular to the river, using the same 120 m of fencing and the same y≤45 constraint. Let x be the depth perpendicular to the river and Y the total length along the river. Determine whether this arrangement can enclose more total area than the best single pen.
Read the worked answer
For two adjacent pens sharing a divider perpendicular to the river, let x be the common depth and Y the total length along the river. Fencing is used on two outer widths, the shared divider, and the back length: Y+3x=120, so Y=120-3x. Total area A(x)=xY=120x-3x^2. Constraints: Y≤45 gives x≥25; Y≥20 gives x≤33.33; also x≥10. Domain [25,33.33]. The vertex is x=20, outside the domain, and the parabola opens downward, so A is decreasing on [25,33.33]. Maximum at x=25, Y=45: A=25×45=1125 m². This is less than the best single pen (1687.5 m²). The two-pen arrangement does not enclose more total area under these constraints.