Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed i nthe region enclosed by the graphs of f(x)=18-x^2 and g(x)=2x^2-9.
Is it equal to about 67.67?
I have no idea finding the answer by using max or min. Can anyone please give me some hints?THANKS A LOT!!!
we are looking at two intersecting parabolas, one opening up the other down
find their intersection points:
2x^2 - 9 = 18 - x^2
.
.
x = ± 3
in the region we want the downwards opening curve is ABOVE the upwards curve, so the effective height in our region is 18-x^2 - (2x^2-9)
= -x^2 + 27
and by symmetry we could take the area from 0 to 3, then double it
so area = 2[integral] (-x^2 + 27)dx from 0 to 3
= 2[(-1/3)x^3 + 27x] from 0 to 3
= 2[-9+81 - 0]
= 144
check my arithmetic, I tend to make a lot of silly errors lately.
yes
I totally don't understand this part:
so area = 2[integral] (-x^2 + 27)dx from 0 to 3
= 2[(-1/3)x^3 + 27x] from 0 to 3
= 2[-9+81 - 0]
= 144
I assumed that from the level of the question you posted, that you would be familiar with how integration is related to finding the area of curved regions.
Are you in a classroom setting of learning Calculus, or are you on your own?
in class
what part confused you?
The fact that I only went from 0 to 3, then doubled?
We could have gone from -3 to 3, let's try that.
so area = [integral] (-x^2 + 27)dx from -3 to 3
= (-9+81)-(9-81)
= (72) - (-72)
= 144
yeah i have a hard time remembering all the symbols but from what i see you managed to understand it just fine.
the integral is a symbol of antiderivative which you managed to do (don't forget the c when you evaluate w/o determining values) then you evaluated antideriv of A - antideriv of B yielding your final result. the area under the curve