Answer:I can help for y= 4-x^2
Step-by-step explanation:
I can help for y= 4-x^2
The dimension of the rectangle would be length=4√3 and width =4-43=83
Explanation:Let one vertex on the x axis be x, then the other vertex would be -x. Length of the rectangle would be thus 2x and width would be 4−x2
Area of the rectangle would be A=2x(4−x2)=8x−2x3
For maximum area dAdx=0=8−6x2, which gives x=2√3
The dimension of the rectangle would be length=4√3 and width =4-43=83
Answer:
(2/3)√15 wide by 10/3 tall, approx 2.582 by 3.333
Step-by-step explanation:
You want the dimensions of the largest rectangle that fits under the graph of y = 5 -x² above the x-axis.
The graph of y=5-x² is symmetrical about the y-axis, so the width of it will be x -(-x) = 2x and the height will be 5-x².
The area will be the product of the width and height:
A = WH
A = (2x)(5 -x²) = 10x -2x³
The area will be a maximum where the derivative of the area function is zero.
dA/dx = 10 -6x² = 0
x² = 10/6
x = (√15)/3
The corresponding rectangle dimensions are ...
width = 2x = (2/3)√15
height = 5 -x² = 5 -5/3 = 10/3
The rectangle with largest area is (2/3)√15 wide by 10/3 tall.
