Question

Find the dimensions of the rectangle of maximum area having two 

vertices on the x-axis and 2 vertices above the x-axis on the graph of y 

= 4 – x2

Answer 1

GIVEN:

According to the given information and symmetry of rectangle

let, vertices of rectangle be,

on x-axis as $\implies (x,0)and(-x,0)$

on the curve $\implies (x,4-x^2) and (-x,4-x^2)$

Therefore;

length of rectangle will be $=2x$

width rectangle will be$=4-x^2$

$\therefore$ area of rectangle $(A)=2x(4-x^2)$

For area to be maximum;

$\boxed{{dA\over dx}=0}\\ \implies8-6x^2=0 \\\implies x={2\over\sqrt3}$

therefore the dimensions of the rectangles will be :

length$=2x={8\over\sqrt3} units$

width$=4-x^2=4-{4\over3}={8\over3}units$

and area=${64\over3\sqrt3}unit^2$