Question

Please help me with this calculus problem!



A rectangle lies in the first quadrant with one vertex at the origin and two of the sides along the co-ordinate axis. The fourth vertex lies on the parabola y = 27-x^2. Find the maximum area of the rectangle and explain why it is maximum.

Answer 1

A(x) = y*x = (27-x^2) *x

Answer 2

if my picture is correct, area of your rectangle is
\[A(x)=x(27-x^2)=27x-x^3\] you want the max on the interval \((0,\infty)\)
take derivative, set it equal to zero to find the critical points and check

Answer 3

What does finding the critical points do? What are critical points?

Answer 4

@Satellite73?

Answer 5

take first derivative, equate it to zero, solve for x ==> those points will be your critical points

Answer 6

\[-3x ^{2}+27=0\]

Answer 7

\[27-3x^2=0\]
\[3(9-x^2)=0\]
\[x=\pm3\] but since you are in quadrant 1 stick with 3

Answer 8

you know it is a local max because
1) you are not an idiot, it has to be a max and cannot be a min
2) you know what a cubic function with negative leading coefficients looks like
3) the derivative changes sign from positive to negateve at \(x=3\) so it must be a max
4) the second derivative is \(-6x\) which is negative at \(x=3\) so your function is concave down at that point and therefore \(x=3\) is a local max

Answer 9

Ok, so how do we find the area?

Answer 10

@satellite73?

Answer 11

put the value 3 on parabolic function, get y and find the area.

Answer 12

your base is \(x=3\) so your height is \(27-3^2=27-9=18\) and your area is whatever the product is

Answer 13

Product?

Answer 14

eighteen times three

Answer 15

LOL

Answer 16

@Satellite73 Thank you very much and sorry for my stupidity :)

Answer 17

no no it is not stupidity. this stuff is hard the first time you see it
yw