Question 823745: A rectangle has one vertex in quadrant I on the graph of y=10-x^2, another at the origin, one on the positive x-axis, and one on the positive y-axis.
a.) Express the area A of the rectangle as a function of x.
b.) Find the largest area A that can be enclosed by the rectangle?
Found 2 solutions by TimothyLamb, KMST:
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
y = 10 - x^2
y = -x^2 + 10
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it is well known that the area of a rectangle is maximized when its sides are of equal length, in other words, when the rectangle is a square
as such:
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y = x = -x^2 + 10
-x^2 - x + 10 = 0
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the above quadratic equation is in standard form, with a=-1, b=-1, and c=10
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-1 -1 10
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has two real roots (the x-intercepts), which are:
x = -3.70156212
x = 2.70156212
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negative length doesn't make sense for this problem, so use the positive root:
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x = length = 2.70156212 units
y = width = 2.70156212 units
NOTE: the problem statement does not specify units, but these could be any linear unit, such as m, cm, ft, etc.
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answerA:
express the area A of the rectangle as a function of x:
A(x) = x * (-x^2 + 10)
A(x) = -x^3 + 10x
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answerB:
find the largest area A that can be enclosed by the rectangle:
A = maximum area = (2.70156212)*(2.70156212)
A = maximum area = 7.298437888218896 sq.units
NOTE: the problem statement does not specify units, but these could be any square unit, such as m^2, cm^2, ft^2, etc.
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Solve quadratic equations, quadratic formula:
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Solve systems of linear equations up to 6-equations 6-variables:
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Answer by KMST(5328)  (Show Source):
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