Answer:
49 and 98, with an area of 4802
Step-by-step explanation:
So, let X be one of the sides, and Y be the other
X*Y would be the area of the rectangle since width*length = area, and X is the width and Y is the length
2X+Y would be the perimeter, excluding the wall (since we don't need fencing for the side sticking to the wall)
To find the max area, we can plot our two equations
so, our two equations:
1. 2X+Y=196 (since perimeter)
2. X*Y=area, and we're trying to maximise it
We can re-arrange the first equation to get
3. Y=196-2X
plug our new equation (#3) into the second equation(#2)
to get:
X*(196-2X)=area=196X-[tex]2X^2[/tex]
And now that we have this equation, we can either:
Graph it to see the largest area (the highest point)
Or calculate the max area using calc
We have a quadratic function that goes up, then down as X increases, so we'll try to find the point where the function stops going up, and instead goes down as X increases (image is attached, not to scale). Recall that the first derivative basically tells is the slope of a function/equation at a given X, so when the function is at a slope of 0, we know that the area is at its max point.
Take the first derivative
(196X-2X^2)'=196-2(2X)=196-4X
196-4X is the equation of the slope, so when X = 49, then the slope is 0
This means that one of the side lengths should be 49
To find out which one, we can use trial and error: (there are 2 cases)
1. 49*((196-49)/2)=3601.5
or:
2. 49*(196-2*49)=4802
The second option is correct since the number, or the area is larger.
Side length 1: 49
Side length 2: (196-2*49), or 98
Note: there's definitely another way to do this problem
