Consider the region in the  rst quadrant whose

Vecna Andres

Vecna Andres

Answered question

2022-06-20

Consider the region in the  rst quadrant whose top portion is from the graph of y = 2 􀀀 x2 while its bottom
portion is from the graph of y = x. A rectangle is to be insribed in the region such that one side of the rectangle
is on the y-axis. Find the area of the largest rectangle that can be inscribed in the region.

Answer & Explanation

star233

star233

Skilled2023-05-21Added 403 answers

To find the area of the largest rectangle that can be inscribed in the given region, we need to analyze the properties of the region and determine the dimensions of the rectangle.
The region in the first quadrant is bounded by two curves: the top portion is given by y=2x2, and the bottom portion is given by y=x. Let's sketch these curves to visualize the region.
First, let's find the intersection points of the curves:
2x2=x
Bringing all terms to one side, we have:
2x2x=0
Factoring out an x, we get:
x(2x1)=0
This equation is satisfied when x=0 or x=12.
Therefore, the curves y=2x2 and y=x intersect at the points (0,0) and (12,12).
Now, let's consider a rectangle inscribed in this region, with one side on the y-axis. Let the height of the rectangle be x (since it is along the y-axis), and the width be y. We need to determine the values of x and y that maximize the area of the rectangle.
The area of the rectangle is given by A=xy.
Since one side of the rectangle is on the y-axis, the width of the rectangle, y, will be the distance between the two curves, which is 2x2x.
Substituting y=2x2x into the area formula, we have:
A=x(2x2x)
Simplifying, we get:
A=2x3x2
To find the maximum area, we can take the derivative of A with respect to x and set it equal to zero:
dAdx=6x22x=0
Factoring out 2x, we have:
2x(3x1)=0
This equation is satisfied when x=0 or x=13.
To determine whether these values correspond to a maximum or minimum, we can take the second derivative of A:
d2Adx2=12x2
Substituting x=0 and x=13, we get:
d2Adx2|x=0=2
d2Adx2|x=13=2
Since d2Adx2|x=13 is positive, it indicates a local minimum. However, since d2Adx2|x=0 is negative, it suggests a local maximum.
Therefore, the maximum area of the rectangle can be achieved at x=13.
Substituting this value back into y=2x2x, we get:
y=2(13)213=13
Thus, the dimensions of the largest rectangle that can be inscribed in the region are x=13 and y=13.
Finally, we can calculate the area of this rectangle:
A=xy=13·13=19.
Hence, the area of the largest rectangle that can be inscribed in the given region is 19.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Multivariable calculus

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?