Area Under the Curve
Area Under Curves
nta_pyq_2025_apr
Grade 12
Question:
Consider the region $R = \left\{(x,y): x\leq y\leq 9-\dfrac{11}{3}x^2,\; x\geq 0\right\}$. The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in $R$, is:
$\dfrac{730}{119}$
$\dfrac{625}{111}$
$\dfrac{821}{123}$
$\dfrac{567}{121}$
Step-by-Step Solution
Key Concept: A rectangle with lower-left corner at $(t,t)$ (on $y=x$) and upper-right corner at $(t, 9-\tfrac{11}{3}t^2)$ has area $A(t) = t\cdot(9-\tfrac{11}{3}t^2-t)$; differentiate and set $A'(t)=0$.
For a rectangle with base on $y=x$: width $= t$, height $= 9-\tfrac{11}{3}t^2-t$.
$A(t) = t\left(9-\tfrac{11}{3}t^2-t\right) = 9t-t^2-\tfrac{11}{3}t^3.$
$A'(t) = 9-2t-11t^2 = 0 \Rightarrow 11t^2+2t-9 = 0 \Rightarrow t = \dfrac{9}{11}$ (taking positive root).
$$A\!\left(\frac{9}{11}\right) = \frac{9}{11}\cdot\frac{63}{11} = \frac{567}{121}.$$
Correct Answer: 4