The area of a rektangel

[Sara33]

A rektangel has two corners at the x-axis and two corners at the curve y=15-x^2. Which x value should the function have if you want to create the greatest area of the rektangel?

First I drew the graph of the function and marked two points on the x-axis. The first point was at the left hand corner of the rektangel and has the coordinates (-a,0). The other point at the x-axis is at the right corner of the rektangel and the coordinates are (a,0). At the left hand side the left corner touches the curve and the coordinates are (-a, 15-a^2). I do not understand why it should be 15-a^2. Can you please explain? At the right side the right corner toches the cruve and has the coordinates (a, 15-a^2).

The base is 2 a (a+a=2a) and the height is 15-a^2 (15-a^2-0=15-a^2). An expression of the area is (2a) (15-a^2). When I solve it,I set the it equal the cero and I get the answer a=3.87. I believe the vertex should be at the point 1.94, however that is incorrect. How should I solve it? Thank you in advance!

 

[wjm11]

A rektangel has two corners at the x-axis and two corners at the curve y=15-x^2. Which x value should the function have if you want to create the greatest area of the rektangel?

First I drew the graph of the function and marked two points on the x-axis. The first point was at the left hand corner of the rektangel and has the coordinates (-a,0). The other point at the x-axis is at the right corner of the rektangel and the coordinates are (a,0). At the left hand side the left corner touches the curve and the coordinates are (-a, 15-a^2). I do not understand why it should be 15-a^2. Can you please explain? At the right side the right corner toches the cruve and has the coordinates (a, 15-a^2).

The base is 2 a (a+a=2a) and the height is 15-a^2 (15-a^2-0=15-a^2). An expression of the area is (2a) (15-a^2). When I solve it,I set the it equal the cero and I get the answer a=3.87. I believe the vertex should be at the point 1.94

In the future, please show *how* you solved it. This is very important, so we can determine where you may have gone wrong. Exactly what was it that you set equal to zero? It is the derivative that must be set to zero.

Since the rectangles are symmetrical on either side of the y-axis, we can consider just the area of the rectangle in the first quadrant: A = a(15 – a^2) and determine when it reaches a maximum. Recognizing that a = x, we can rewrite this as A = x(15 – x^2). Expanding this we get A = 15x – x^3. To determine maximum and minimum values of this function, we find the derivative with respect to x, and set it equal to zero:

A = 15x – x^3
A’ = 15 – 3x^2
0 = 15 – 3x^2
-15 = -3x^2
5 = x^2
x = +/- sqrt(5)

Therefore, x = sqrt(5) is the width of the rectangle in the first quadrant. The height is 15 – x^2, so

h = 15 – (sqrt(5))^2 = 15 – 5 = 10

Therefore, the area of the rectangle in the first quadrant is A = (sqrt(5))(10).

The area under the curve in both quadrants I and II is twice this amount, so A = (sqrt(5))(20).

Please check my work for accuracy.

 

[Sara33]

Thank you for your help! I have heard about derivative, however I do not know how to calculate it since derivative is included in my math course next term. I believe this is beyond by knowledge for know. I did this exercisse since I found it in an advanced section in my book. I will take a look at your answer again in a few weeks when I will learn about derivative.