Answer to Question #194624 in Calculus for Moe

Question

Find the largest possible area of a shaded rectangle which lies within parabola $g(x) = - x^2 + 12$ which has its base on the x-axis and two of its vertices lying on the parabola and above the x-axis.

Answer 1

Question:-

Find the largest possible area of a shaded rectangle which lies within parabola $g(x) = - x^2 + 12$ which has its base on the x-axis and two of its vertices lying on the parabola and above the x-axis.

Solution:-

according to question,

we get diagram,

such that as image

here,

coordinate of A and B ,assume (-x,0) and (x,0) respectively.

and

coordinate of C and D will be

(x,(-x²+12)) and (-x,(-x²+12)) respectively, becouse y coordinate given as (-x²+12) in the question.

now we find largest possible area of rectangle.

here in image,

rectangle lenght=AB=2x

rectangle width=CD=(-x²+12)

So

rectangle area=2x.(-x²+12)

let it is a function of x, such that

f(x)=2x.(-x²+12)

now we want largest value of this,

so we will maxima this,

so we find f'(x),

f'(x)=2(-x²+12)+2x(-2x)

now put f'(x)=0

2(-x²+12)+2x(-2x)=0

-6x²+24=0

6x²=24

x=±2

now put positive value of x in the f(x)

and we will be get largest area of rectengle.

so put

f(2)=2(2)(-2²+12)

=32

largest possible area of a rectangle= 32