The georectangle problem: area as a parabola

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In a fairy-tale land a “georectangle” has an area described by a parabola depending on $x$ (its age in years). The georectangle is born with area $0$ at $x=0$; at age $4$ its area is $8$; at age $10$ its area is $5$.

• (a) find how the area depends on $x$ and describe its graph;
• (b) find at what age the area returns to $0$ (it dies);
• (c) find at what age the area is maximum (it is a square).

Solution

(a) Model. Write the area as $A(x)=ax^2+bx+c$. Passing through $(0;0)$ gives $c=0$. The remaining conditions are: $A(4)=8:\ 16a+4b=8 \Rightarrow 4a+b=2$ $A(10)=5:\ 100a+10b=5 \Rightarrow 10a+b=0.5$ Subtracting the first from the second: $6a=-1.5 \Rightarrow a=-\dfrac14$, hence $b=2-4a=2+1=3$. $A(x)=-\frac14x^2+3x$ It is a parabola opening downward (since $a<0$), passing through the origin.

(b) When it dies. $A(x)=0$: $x\left(-\frac{x}{4}+3\right)=0 \Rightarrow x=0 \ \text{or}\ x=12$ After birth, the area returns to $0$ at $x=12$ years.

(c) Maximum area. The vertex has $x_V=-\frac{b}{2a}=-\frac{3}{2\left(-\frac14\right)}=6, \qquad A(6)=-\frac14(36)+3(6)=-9+18=9$ The area is maximum ($9$, when it is a square) at $x=6$ years. $\boxed{A(x)=-\tfrac14x^2+3x\quad \text{dies at }x=12\quad \text{max }A=9\text{ at }x=6}$