You have 132 m of fencing with which to make 2 side by side rectangular enclosures against an existing wall. If the enclosures are adjacent and of the same depth, what is the maximum area that can be enclosed Please answer ASAP will give out brainiest

The two enclosures will need three equal fences coming out from the wall and meeting another fence running parallel to the wall. If the fences coming out from the wall are x metres long the parallel fence will be (132 - 3x) metres long.
The area A = x(132 - 3x) = 132x - 3x^2
The derivative of A = zero when 132 - 6x = 0 which means the maximum area is when x = 22m
The maximum area = 22 x (132 - 3 x 22) = 1452 m^2

If you don’t know how to find derivatives then you could sketch the graph of y = x(132 - 3x).
This is an inverted parabola (hill) with x intercepts at 0 and 132/3 = 44.
The maximum point (top of the hill) is halfway between 0 and 44 I.e. 22m

Try any other value for x and the area will be smaller.

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph

there is a hike in both the graphs so the graphs are increasing on their domain.x-intercept of log x and log 6x are differentthere is a vertical asymptote of both the graphs so there is no y-interceptas both the graphs are increasing so there is no horizontal asymptote