What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 12 - x2? (4 points)

16 square units

Step-by-step explanation:

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(12 - x²)  

= -x³ + 12x  

d(area)/dx = 3x² - 12 = 0 for a maximum of area  

3x² = 12  

x² = 4  

x = ±2  

y = 12-4 = 8

So, the largest area = 2 x 8 = 16 square units

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25

(-5,-28)

i graphed to on desmos.