Download The area of the rectangle is given by (2x)(2y). We can

Calculus I (MAC2311-v1)
Quiz 5 (2014/11/04)
Name (PRINT):
An answer with no work or reasoning receives no credit.
You may not use any calculators.
1. (7 points) Find the dimensions of the rectangle of largest area that can be inscribed in
a circle of radius r. (Sketch a diagram with all the necessary quantities clearly labeled.)
Solution:
The area of the rectangle is given by (2x)(2y). We can write y in terms of x and r
r 2 = x2 + y 2 ,
√
y = r 2 − x2 .
Now, the area can be written as
√
A(x) = (2x)(2 r2 − x2 )
√
= 4x r2 − x2 .
Geometrically, A(x) is defined only when x is in D = [0, r]. At the end points the
rectangle degenerates into a line.
Differentiating we get
√
−2x
A0 (x) = 4 r2 − x2 + 4x √
2 r 2 − x2
√
4x2
= 4 r 2 − x2 − √
.
r 2 − x2
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Setting A0 (x) to zero gives
√
4x2
4 r 2 − x2 = √
,
r 2 − x2
r 2 − x2 = x2 ,
2x2 = r2 ,
which in the domain D gives x =
√r ,
2
our first critical point.
Other critical points come from the points at which A0 (x) is not defined:
r2 − x2 = 0,
x = r.
Evaluting A at the end points of the domain and and at the critical points we get
A(0) = 0,
A(r) = 0,
r
r
r
r
A( √ ) = 4( √ ) r2 − ( √ )2
2
2
2
2
= 2r .
So, dimensions of the rectangle of largest area that can be inscribed in a circle of
radius r are
√
r
width = 2x = 2( √ ) = 2r,
2
r
√
r
height = 2y = 2 r2 − ( √ )2 = 2r.
2
In other words, the optimal rectangle (in the sense above) is a square. (This may
have been expected from symmetry.)
2. (3 points) Using Newton’s method and starting from the initial approximation of x1 = 0,
find x2 , the second approximation to the root of the equation x − e−x = 0.
Solution: Put f (x) = x − e−x . Differentiating we get f 0 (x) = 1 + e−x . At x1 = 0
we have f (x1 ) = −1 and f 0 (x1 ) = 2.
Starting Newton’s iteration
xn+1 = xn −
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f (xn )
f 0 (xn )
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from x1 = 0 we get
f (x1 )
f 0 (x1 )
−1
=0−
2
1
= .
2
x2 = x1 −
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