SOLUTION: A rectangular box is to be placed in the first quadrant in such a way that one side lies on the positive x-axis and one side lies on the positive y-axis. The box is to lie below th

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Question 118436: A rectangular box is to be placed in the first quadrant in such a way that one side lies on the positive x-axis and one side lies on the positive y-axis. The box is to lie below the line y=-2x+5. Give the dimensions of such a box having greatest possible area.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given:
y=+-2x%2B5
known: positions for the rectangular box defined by the x and y minimums and maximums
so, to find out what the dimensions of such a box are, first find midpoint coordinates of the line segment between x-intecept and y-intecept or
between points
x%5B1%5D,y%5B1%5D= 2.5,0 and
x%5B2%5D,y%5B2%5D= 0,5
That will be:
x coordinate of mid point is %28x%5B1%5D+%2B+x%5B2%5D%29%2F2+=+%282.5+%2B+0%29%2F2=1.25
y coordinate of mid point is %28y%5B1%5D+%2B+y%5B2%5D%29%2F2+=+%285+%2B+0%29%2F2=2.5

The mid point of segment joining two point is:
(1.25,2.5)


Here is the graph that shows the point (1.25,2.5):

Solved by pluggable solver: FIND EQUATION of straight line given 2 points
hahaWe are trying to find equation of form y=ax+b, where a is slope, and b is intercept, which passes through points (x1, y1) = (1.25, 2.5) and (x2, y2) = (0, 5).
Slope a is a+=+%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29+=+%285-2.5%29%2F%280-1.25%29+=+-2.
Intercept is found from equation a%2Ax%5B1%5D%2Bb+=+y%5B1%5D, or -2%2A1.25+%2Bb+=+5. From that,
intercept b is b=y%5B1%5D-a%2Ax%5B1%5D, or b=2.5--2%2A1.25+=+5.

y=(-2)x + (5)

Your graph:






the dimensions of a box are: x+=1.25 and y+=+2.5

its greatest possible area is:
A+=+x%2Ay+
A+=+1.25%2A2.5
A+=+3.125