Respuesta :
The answer is 54 square units.
let the vertex in quadrant I be (x,y)
then the vertex in quadratnt II is (-x,y)
base of the rectangle = 2x
height of the rectangle = y
Area = xy
= x(27 - x²)
= -x³ + 27x
d(area)/dx = 3x² - 27 = 0 for a maximum of area
3x² = 3 x 3² = 27
x² = 9
x = ±3
y = 27-9 = 18
So, the largest area = 3 x 18 = 54 square units
let the vertex in quadrant I be (x,y)
then the vertex in quadratnt II is (-x,y)
base of the rectangle = 2x
height of the rectangle = y
Area = xy
= x(27 - x²)
= -x³ + 27x
d(area)/dx = 3x² - 27 = 0 for a maximum of area
3x² = 3 x 3² = 27
x² = 9
x = ±3
y = 27-9 = 18
So, the largest area = 3 x 18 = 54 square units
Answer:
108 units squared
Step-by-step explanation:
Length is 2x , width is y
area = 2x * y
sub for y
area = 2x(27 - x^2)
area = 54x - 2x^3
1st derivative: 54 - 6x^2
54 - 6x^2 = 0
6(9 - x^2) = 0
6(3 - x)(3 + x) = 0
so, x = 3
solve for y: 27 - 3^2 = 18
area = (3*2)* 18 = 108 units squared
