Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 11 − x 2 . What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

Answer:$\frac{22}{3}$,$2\dot \sqrt{\frac{11}{3}}$

Step-by-step explanation:

Given

rectangle with its base on x-axis

and other two corners at parabola

and parabola is downward facing symmetric about y-axis

let y be the y co-ordinate of the corner thus x co-ordinate is given by

$x=\pm \sqrt{11-y}$

Thus lengths of rectangle is $2\sqrt{11-y}$ & y

Area $=y\times 2\sqrt{11-y}$

differentiating w.r.t to y for maximum area

$\frac{\mathrm{d} A}{\mathrm{d} y}=2\times \sqrt{11-y}-\frac{y}{2\dot \sqrt{11-y}}=0$

we get y=$\frac{22}{3}$

and $x=\pm \sqrt{\frac{11}{3}}$

A_{max}=16.21 units