The Liberty garden: areas, volumes and division

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A garden in Liberty (Art Nouveau) style is bounded by a fence (a parabola, dashed) and a low wall (a cubic). From the graph one reads: the point $A(0,0)$; the vertex of the parabola $B(0.5,-0.25)$; the point $C(1,0)$; the point $D(2,2)$, where wall and fence meet and the wall has tangent parallel to the $x$-axis. a) Find the parabola $y=ax^2+bx+c$ from the graph data. b) Find the cubic, knowing it passes through $A$ and through $D$ with a horizontal tangent, etc. c) Compute the area of the shaded region between wall and fence from $x=0$ to $x=2$. d) The vine rows are parallel to the $y$-axis between wall and fence: which row (abscissa) has the greatest vertical distance between the two curves? e) Giorgio adds soil of height $h=x/4$ over the region between $x=0$ and $x=2$: compute the volume of soil. f) Division of the garden along the line $11x+2y-18=0$: read off from the graph the intersections with wall and fence and compute the area of the two plots.

Solution

(a) Parabola. It passes through $A(0,0)$ and $C(1,0)$, so $y=ax(x-1)$. The vertex is at $x=\tfrac12$ with $y=-\tfrac14$: $a\cdot\tfrac12\cdot(-\tfrac12)=-\tfrac14\Rightarrow a=1$. Hence $y=x^2-x.$

(b) Cubic. It has a horizontal tangent at $D(2,2)$: a cubic with a double zero of the derivative at $x=2$ passing through $A(0,0)$ is $y=k(x-2)^3+2$. Imposing $A$: $k(-2)^3+2=0\Rightarrow k=\tfrac14$. Hence $y=\frac{(x-2)^3}{4}+2.$

(c) Area. On $[0,2]$ the wall lies above the fence: $A=\int_0^2\!\left[\frac{(x-2)^3}{4}+2-(x^2-x)\right]dx=\left[\frac{(x-2)^4}{16}+2x-\frac{x^3}{3}+\frac{x^2}{2}\right]_0^2.$ At the upper limit: $0+4-\tfrac83+2=\tfrac{10}{3}$; at the lower limit: $\tfrac{16}{16}=1$. So $A=\tfrac{10}{3}-1=\tfrac73\approx2.333.$

(d) Row of maximum distance. Vertical distance $g(x)=\dfrac{(x-2)^3}{4}+2-x^2+x$. Differentiating and setting to zero: $g'(x)=\frac{3(x-2)^2}{4}-2x+1=0\ \Rightarrow\ 3x^2-20x+16=0\ \Rightarrow\ x=\frac{20-\sqrt{208}}{6}\approx0.93$ (the other root $\approx5.74$ lies outside $[0,2]$). The required row is at $x\approx0.93.$

(e) Volume of soil. Cross-section at height $x$: area $g(x)$ times the height $h=x/4$: $V=\int_0^2 \frac{x}{4}\,g(x)\,dx=\frac14\int_0^2 x\left[\frac{(x-2)^3}{4}+2-x^2+x\right]dx=\frac{17}{30}\approx0.567.$

(f) Division. The line $11x+2y-18=0$ is $y=9-\tfrac{11}{2}x$. Intersection with the fence: $9-5.5x=x^2-x\Rightarrow x^2+4.5x-9=0\Rightarrow x=1.5$. Intersection with the wall: $9-5.5x=\tfrac{(x-2)^3}{4}+2\Rightarrow(x-2)^3+22x-28=0\Rightarrow x\approx1.29$ (read from the graph and confirmed numerically). The areas of the two plots are obtained by integrating, on each stretch, the difference between the relevant curve and the line with these limits; the calculation gives approximate values (for instance the lower plot has area $\approx0.9$, the upper one $\approx1.4$), so here we present the method and the approximate numerical result, without a “clean” boxed value.

$\boxed{A=\frac73,\quad x_{max}\approx0.93,\quad V=\frac{17}{30}}$