Answer:
D
Step-by-step explanation:
x/(x²+3x+2) - 1/[(x+2)(x+1)]
x² + 3x + 2 = x² + 2x + x + 2
= x(x + 2) + (x + 2)
= (x + 2)(x + 1)
= x/[(x+2)(x+1)] - 1/[(x+2)(x+1)]
= (x-1)/[(x+2)(x+1)]
= (x-1)/(x²+3x+2)
F(x) = (x + 1)(x - 2)
f(x) = x(x - 2) + 1(x - 2)
f(x) = x(x) - x(2) + 1(x) - 1(2)
f(x) = x² - 2x + x - 2
f(x) = x² - x - 2
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.
2(x+y) = 300
x+y = 150
y = 150-x
A=x(150-x) <--(substitution)
The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150
So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.
A=75(150-75)
A=75*75
A=5625
So the maximum area that can be enclosed is 5625 square feet.
