Find the maximum area of a rectangle that is inside of the triangle forms by the x-axis and the lines y=-3x+12 and y=3x+12. The base of the triangle is on the x-axis and the two upper verticies are on the lines y=-3x+12 and y=3x+12.
With this question. Obviously the are of the rectangle is lw or xy… I have figured out that x = (8-2x) and y=(12-y). Im not sure of the next step.. Could someone help?
Best Answer
The two lines are symmetric across the $y$-axis. Therefore, just take twice the area in the 1st quadrant.
Thus the area of the rectangle (width times height) is $2x(-3x+12) = -6x^2+24x$, where $0 ≤ x ≤ 4$.
Since the quadratic is concave up, notice that the maximum is at $x = -\frac{b}{2a} = -\frac{24}{2(-6)} = 2$, which gives the maximum area as $-6(2)^2 + 24(2) = \boxed{24}$.