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.
So, when x = -10.
If you notice, -10 is NOT in the domain (the domain is (-3,13))
so, there is no solution.
Answer:
78
Step-by-step explanation:
the product of 6 and 13 =78
When the x-intercept solution of a system of two-variable equations is plugged in, the equations yield the same value. Thus, for the points to be solutions, F(x) = G(x)
The only option that shows this is C, where
F(8) = G(8) and F(24) = G(24)
