A rectangle is inscribed in the interior section cut from the parabola x = 1/4 y2 by the line x=3. One side of the rectangle lies on the line. Fine the maximum area of such a rectangle.

Make a sketch.

Let the points on the parabola be P(x,y) and Q(x,-y)

Area = 2y(3-x)
= 2y(3-(1/4)y^2)
= 6y - (1/2)y^2
d(Area)/dy = 6 - y = 0 for a max of Area
y = 6

the max area is 6(6) - (1/2)(36)
= 18 square units

check my arithmetic

To find the maximum area of the rectangle inscribed in the given parabola, we need to determine the dimensions of the rectangle first.

Let's start by visualizing the problem. We have the parabola given by the equation x = (1/4)y^2 and the line x = 3. We want to inscribe a rectangle in the region between these two curves.

First, we need to find the points of intersection between the parabola and the line. To do this, we'll substitute x = 3 into the equation of the parabola:

3 = (1/4)y^2

Multiply both sides by 4 to get rid of the fraction:

12 = y^2

Taking the square root of both sides, we find:

y = ±√12

Since we are looking for the positive y-values, we have y = √12 = 2√3.

Now, let's find the corresponding x-values for y = 2√3:

x = (1/4)(2√3)^2
= (1/4)(4*3)
= 3

Therefore, the two points of intersection are (3, 2√3) and (3, -2√3). We can see that one side of the rectangle lies on the line x = 3.

Next, we need to find the width of the rectangle. Since one side of the rectangle is on the line x = 3, the width of the rectangle is the difference between the y-values of the two points of intersection:

Width = 2√3 - (-2√3)
= 2√3 + 2√3
= 4√3

Finally, we need to find the height of the rectangle. The height is the distance between the parabola and the x-axis at the point (3, 2√3). We can do this by finding the y-coordinate of the point on the parabola with x = 3. Substitute x = 3 into the equation of the parabola:

x = (1/4)y^2
3 = (1/4)y^2

Multiply both sides by 4:

12 = y^2

Taking the square root of both sides, we find:

y = ±√12

Again, since we are looking for the positive y-value, we have y = √12 = 2√3.

Therefore, the height of the rectangle is 2√3.

Now that we have the width and height of the rectangle, we can calculate its area:

Area = Width * Height
= (4√3)(2√3)
= 8(√3)^2
= 8 * 3
= 24

So, the maximum area of the rectangle inscribed in the given parabola is 24 square units.