Question

Consider the collection of all curve of the form $y=a-b{x}^2$ that pass through the point $(2,1)$, where $a$ and $b$ are positive constant. Determine the value of $a$ and $b$ that will minimise, the area of the region bounded by $y=a-b{x}^2$ and $x$-axis. Also find the minimum area

• A. $b=1/8,A_{minimum}=4\sqrt{3}sq.units$
• B. $b=1/4,A_{minimum}=4\sqrt{3}sq.units$
• C. $b=1/4,A_{minimum}=2\sqrt{3}sq.units$
• D. $b=1/8,A_{minimum}=2\sqrt{3}sq.units$

Answer 1

Answer: (A)

$b=1/8,A_{minimum}=4\sqrt{3}sq.units$

Curve $y=a-b{x}^2$ passes through the point $(2,1)$
$\therefore a-4b=1$
$A=2{\int }_0^{\sqrt{a/b}}(a-b{x}^2)dx=2{[ax-\frac{b{x}^3}{3}]}_0^{\sqrt{a/b}}$
$=\frac{4}{3}\frac{a^{3/2}}{\sqrt{b}}=\frac{4}{3}\frac{{(1+4b)}^{3/2}}{\sqrt{b}}$

$\Rightarrow A^{\prime }=\frac{2}{3}\frac{\sqrt{1+4b}(8b-1)}{b^{3/2}}$

$\Rightarrow A^{\prime }=0$

$\Rightarrow b=\frac{1}{8}$
$\Rightarrow A=4\sqrt{3}sq.units$