SOLUTION: Q: Find the area A of the largest rectangle with base on the x-asix that can be inscribed in the region R bounded above the by the growth of y = 9 -x^2 and below by the x-axis.
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Question 1001517: Q: Find the area A of the largest rectangle with base on the x-asix that can be inscribed in the region R bounded above the by the growth of y = 9 -x^2 and below by the x-axis.
A:I know that y = -x^2 + 9 is an inverted parabola that is shifted upwards 9 units because + 9 and hase x points on -3 and +3.
I know that the area of a rectangle is A=L*W or in this case I made it A=x*y for the x and y axis. So the x base stretches from both the [-3,3] and the y goes from [0,9] what I don't get is why is the Area set up like A=2x*y because I know we have the top and bottom part of the rectangle for X so 2X and the left and right side of the rectangle 2Y so shouldn't we take into account that there are two sides of Y representing the left and right side of the rectangle?
But it's not for some reason basic reason I am sure that I forgot. I think I might be confusing permiter or something. I get that confused a lot.
I think the answer goes:
A = 2x*y
A = 2x*(-x^2+9)
A = -2x^2 + 18x
then optimizing it
A' = -4x + 18
find the critical value of x
x=-/+√(9/2) but since we are talking about distance it's always the pos end.
so,
x=+√(9/2)
then plugging this back into my base form before the derivative:
A = - 2(√(9/2)^2 + 18(√(9/2)
Is this correct? what is my issue with the Area formula?
Thank you Answer by Theo(13342) (Show Source):
I think the answer goes:
A = 2x*y
A = 2x*(-x^2+9)
A = -2x^2 + 18x ****************************************
A = -2x^2 + 18x is where you went wrong.
2x * (-x^2 + 9) is equal to -2x^3 + 18x.
find the derivative of that and you get -6x^2 + 18.
set that equal to 0 and you get -6x^2 + 18 = 0
add 6x^2 to both sides of that and you get 6x^2 = 18
divide both sides of that by 6 and you get x^2 = 3
take the square root of that and you get x = plus or minus sqrt(3).
the value of x is plus or minus sqrt(3).
the height of your rectangle is y = 9-x^2
the width of your rectangle is 2x.
the maximum area is when x = plus or minus sqrt(3).
when x = sqrt(3), your area is (9 - sqrt(3)^2) * 2 * sqrt(3)) which is equal to 20.78460969.
you can also find this graphically by graphing y = (9 - x^2) * 2 * x.
the graph of that equation is shown below:
that's the graph of the area of your rectangle and it shows that the maximum area is when x = 1.732 which is the rounded version of sqrt(3).
