A rectangle has a vertex on the line 3x + 4y = 12.
Two of its sides lie along the positive x and y axes.
Find the maximum possible area of such a rectangle.
To find the maximum possible area of the rectangle, we need to first determine the coordinates of the vertices of the rectangle.
Let's start by finding the vertex of the rectangle on the line 3x + 4y = 12. Rearranging the equation in terms of x, we have:
3x = 12 - 4y
x = (12 - 4y) / 3
Since one of the sides of the rectangle lies on the x-axis, the y-coordinate of this vertex will be 0. Substituting y = 0 into the equation, we can find the x-coordinate:
x = (12 - 4 * 0) / 3
x = 12 / 3
x = 4
So, one vertex of the rectangle is (4, 0).
Similarly, since the other side lies on the y-axis, the x-coordinate of the second vertex will be 0. Substituting x = 0 into the equation, we can find the y-coordinate:
3 * 0 + 4y = 12
4y = 12
y = 12 / 4
y = 3
Therefore, the second vertex of the rectangle is (0, 3).
Now, we have the coordinates of two vertices of the rectangle: (4, 0) and (0, 3). To find the maximum possible area, we need to determine the length and width of the rectangle.
The length of the rectangle is the distance between the x-coordinates of the vertices, which is |0 - 4| = 4 units.
The width of the rectangle is the distance between the y-coordinates of the vertices, which is |3 - 0| = 3 units.
Now, we can calculate the area of the rectangle:
Area = length * width
Area = 4 * 3
Area = 12 square units
Therefore, the maximum possible area of the rectangle is 12 square units.
Let the vertex on the line 3x + 4y = 12 be (x,y)
then y = (12-3x)/4 = 3- 3x/4
area = xy = x(3 - 3x/4) = 3x - 3x^2 /4
d(area)/dx = 3 - (3/2)x
= 0 for a max of area
(3/2)x = 3
3x = 6
x = 2 , y = 3 - 6/4 = 3/2
max area = xy = 2(3/2) = 3 units^2