SOLUTION: Hi, I could really use some help with this problem. A rectangle has one vertex on the line {{{y=8-2x}}} x>0, another at the orgin, one on the positive x-axis, and one on the posi

Algebra ->  Algebra  -> Rectangles -> SOLUTION: Hi, I could really use some help with this problem. A rectangle has one vertex on the line {{{y=8-2x}}} x>0, another at the orgin, one on the positive x-axis, and one on the posi      Log On

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Question 76192This question is from textbook College Algebra Esssentials
: Hi, I could really use some help with this problem.
A rectangle has one vertex on the line y=8-2x x>0, another at the orgin, one on the positive x-axis, and one on the positive y-axis.
Draw the graph. Express the area of rectangle A(x) as a function of horizontal side x. Find the dimensions and area of largest rectangle that can fit under the line in Quadrant 1.
Thank you,
JodiThis question is from textbook College Algebra Esssentials

Answer by scott8148(5880) About Me  (Show Source):
You can put this solution on YOUR website!
the area of the rectangle is the product of the x and y coordinates on the y=8-2x line...this means A(x)=x(8-2x)...or A%28x%29=8x-2x%5E2

the graph of A(x) is a downward opening parabola with zeros at x=0 and x=4...the axis of symmetry (and the maximum point) is midway between the zeros at x=2

so for x=2...A(x)=8...and y=4