Let x and y be the dimensions of the rectangle. If the perimeter is 40, we have
![2(x+y)=40 \iff x+y=20](https://tex.z-dn.net/?f=2%28x%2By%29%3D40%20%5Ciff%20x%2By%3D20)
We can expression one variable in terms of the others as
![x+y=20 \iff x=20-y](https://tex.z-dn.net/?f=x%2By%3D20%20%5Ciff%20x%3D20-y)
Since the area is the product of the dimensions, we have
![xy=(20-y)y=-y^2+20y](https://tex.z-dn.net/?f=xy%3D%2820-y%29y%3D-y%5E2%2B20y)
This is a parabola facing down, so it's vertex is the maximum:
![f(y)=-y^2+20y \implies f'(y)=-2y+20](https://tex.z-dn.net/?f=f%28y%29%3D-y%5E2%2B20y%20%5Cimplies%20f%27%28y%29%3D-2y%2B20)
So, the maximum is
![f'(y)=0 \iff -2y+20=0 \iff 2y=20 \iff y=10](https://tex.z-dn.net/?f=f%27%28y%29%3D0%20%5Ciff%20-2y%2B20%3D0%20%5Ciff%202y%3D20%20%5Ciff%20y%3D10)
And since we know that
, we have
as well.
This is actually a well known theorem: out of all the rectangles with given perimeter, the one with the greatest area is the square.
Answer:
y=9
y=1
y= -7
y= -15
Step-by-step explanation:
You have to use the number in the x colum and fill in itnot te equation
y= -4x+1
Start with the first number, -2
y= -4(-2)+1 multiply -4 and -2 because of PEMDAS
y=8+1 simplify
y=9
Now do the next number, 0
y= -4(0)+1 Multiple -4 and 0 because of PEMDAS (Anything times 0 is 0)
y=1
Now do the next number, 2
y= -4(2) + 1 Multiply -4 and 2 because of PEMDAS
y= -8 + 1 Simplify
y=-7
Now do the next number, 4
y= -4(4) +1 Multiply -4 and 4 because of Pemdas
y = -16 +1
y= -15
Answer:
3x (both) 1 and -1
Step-by-step explanation:
Have a good day