Area under a graph problem?

Question

A rectangle is inscribed in the region bounded by the x-aix, the y-axis, and the graph of x-2y-8=0. Write the area as a function of x. Then determine the domain of the function. What are the dimensions of the rectangle that will produce the maximum area?

Please explain the steps to obtain the area as a function of x as well. Thanks!

Answer 1

the line is y = .5x - 4. This line, along with the x-axis and y-axis, form a triangle that's situated below the x-axis. Call the width of the rectangle x, then the height is -y = -.5x + 4. (the reason I used -y is because the value of y is negative due to the region being below the x-axis, but area is always positive, so I used its absolute value). So the area = x(-.5x+4) = -.5x² + 4x.

A(x) = -.5x² + 4x

I may be wrong to assume you're in algebra and not calculus, so I am showing a non-calculus method. As you have probably learned in your class, this is a parabola which has a vertex that represents a maximum since it opens downward. The vertex (h,k) is found with the formula h=-b/2a = -4/-1 = 4. Therefore, the maximum area occurs at A(4) = 8.

The domain of the problem is 0

Answer 2

First, plot the line so you can get and understanding of the problem.

Next, you know the area of a rectangle is going to be the
x coordinate multiplied by the y coordinate, and y is related to x by y = 1/2*x - 4.

Do the multiplication:

x*y = x*(1/2*x - 4) = (x^2)/2 - 4x

This is the basically the answer. However there are two more points that need to be mentioned. First, area cannot be negative and this rectangle lives in the forth quadrant, and second, you need to put the bounds in. So your final answer should look like:

for 0
Taking the derivative of this will show that the maximum area is at x = 4, which is a cogent solution for this type of problem.

Answer 3

Assuming you mean to inscribe the rectangle with one side on the x-axis and one side on the y-axis, one side of the triangle will have length x and the other length 4-(1/2)x. This is because y=(1/2)x-4 and we are interested in values of x between 0 and 8. y will be negative there and since we want our area to be positive we change y's sign. So our area equals x(4-1/2x)=4x-1/2x^2 with a domain from 0 to 8. This is a quadratic with a negative coefficient for x^2, so it is maximized at x=(-4)/(-1)=4. The dimensions of the rectangle are 4 by 2.

Answer 4

ALWAYS, always DRAW A PICTURE OF THE PROBLEM. Here, we have a triangle with vertices of (0,0), (8,0), and (0,-4). This is an area between (x/2)-4=y and y=0 with a domain of 0 to 8. To maximize the area of an inscribed rectangle, the equation is A = x * y, where y is given by the expression above.
Then dA/dx = d(x^2/2 - 4*x)/dx = x-4. Set x-4=0 to get x=4 and y=2 (actually -2, but whose watching) and A = 8

Answer 5

it would be easier if you rotated the graph around if you are confused, or even if you switched x and y, but you would have to remember to switch back for the final answer so, here it goes :) if you rotated the graph, it would be easy to see that the upper function is x=e^y, and the lower function is y^2-2 so set up your integration as ∫(e^y-(y^2-2))dy from -1 to 1 =∫(e^y)dy-∫y^2dy-∫2dy from -1 to 1 =e^y [from -1 to 1] - y^3/3 [from -1 to 1] -2y [from -1 to 1] =(e-1/e)-(1/3-1/3)-(2-(-2)) =e-(1/e)-4