A rectangle is constructed with one side on the positive x-axis, one side on the
ID: 2861935 • Letter: A
Question
A rectangle is constructed with one side on the positive x-axis, one side on the positive y-axis, and the vertex opposite the origin on the line y = 6 - 3x. What dimensions maximize the area of the rectangle? What is the maximum area? Is it possible to construct a rectangle with a greater area than that found in part (a) by placing one side of the rectangle on the line y = 6 - 3x and the two vertices not on that line on the positive x- and y-axes? Find the dimensions of the rectangle of maximum area that can be constructed in this way. A(a) = a (6 - 3a) The maximum area of the rectangle in this situation occurs when the shorter side has length 1 and the longer side has length 3. (Simplify your answers.) The maximum area in this situation is 3. (Simplify your answer.) Let the vertices on the x- and y-axes be (a,0) and (0,b), respectively. Find a function A(a) that gives the area of the rectangle in this situation. Express it entirely in terms of a. A(a) =Explanation / Answer
(a) Area = x y or x(6-3x)
Area = A= x(6-3x)=6x-3x^2
for Maximum Area, dA/dx=6-6x=0
=> x=6/6=1 and y=6-3x=6-3=3
Therefore Maximum Area = x(6-3x)=1(6-3)=3
(b) let the vertices on the x and y axis be (a,0) and (0,b) and other side on the line y=6-3x =>y+3x-6=0
Therefore perpendicular distance from point (a,0) on line y+3x-6=0
is (0+3a+6)/sqrt(1+9)=(3a+6)/sqrt(10).........(1)
Similarly perpendicular distance from point (0,b) on line y+3x-6=0
is (b+0+6)/sqrt(1+9)=(b+6)/sqrt(10)............(2)
From equation 1 and 2
(b+6)/sqrt(10)=(3a+6)/sqrt(10). =>b=3a
Now the distance between (a,0) and (0,b) is sqrt(a^2+b^2)=sqrt(a^2+9a^2)=a* sqrt(10)
Therfore Area=(a* sqrt(10))*((3a+6)/sqrt(10))=a(3a+6)