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.
Answer:
3) 12 yd^2
4) 27.5 m^2
5) 48 ft^2
6) 96 m^2
Step-by-step explanation:
3) 4(2x3x0.5) = 12 yd^2
4) (9x5)-(0.5)(7x5) = 27.5 m^2
5) (10x6)-(6x2) = 48 ft^2
6) 12x8 = 96 m^2
Y = f(3f - 4g + 3) +7g(3f - 4g + 3)
= 3f^2 - 4fg + 3f + 21fg - 28g^2 + 21g
= 3f^2 + 3f + 17fg + 21g - 28g^2
Convert the fractions to decimals:
3/5=0.6
4/5=0.8
You have to find 2 fractions between these 2 numbers. You could pick anything.
7/10= 0.7 = less than 0.8 but greater than 0.6
75/100=0.75 = less than 0.8 but greater than 0.6
Hope this helps