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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Problem Set 8, Part B Solutions -- December 1, 2017}
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\noindent
{\it Section 4.4}
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22.  Assuming the graph is $y = f(x)$, there is just one inflection
point at $x = c$.  The graph of $f(x)$ is concave down on $(0,c)$ and concave
up on $(c,g)$.  
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\noindent
23.  Now assuming the graph is $y = f'(x)$, there are inflection
points for $y = f(x)$ at $x = b$ and $x = e$.  The graph of $f(x)$
is concave up on $(0,b)$ and $(e,g)$ (the intervals where $f'(x)$
is increasing) and the graph of $f(x)$ is concave down on $(b,e)$
(the interval where $f'(x)$ is decreasing).  
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\noindent
24.  Now assuming the graph is $y = f''(x)$, there inflection points
for $y = f(x)$ at the points where $f''(x)$ changes sign, so 
at $x = a,d,f$.  The graph $y = f(x)$ is concave up on the 
intervals where $f''(x) > 0$ (that is, where the graph $y = f''(x)$
is above the $x$-axis), so $(a,d)$ and $(f,g)$.  The graph
$y = f(x)$ is concave down on the intervals where $f''(x) < 0$
(that is, where the graph $y = f''(x)$ is below the $x$-axis), 
so $(0,a)$ and $(d,f)$.  
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\noindent
{\it Section 4.7}
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\noindent
37.  The blue rectangle in the figure is fixed with sides of length $H,L$.
The red rectangle depends on the angle $\theta$ shown, and has area determined
like this.  First look at the triangle at the top.  This is a right triangle
with hypotenuse equal to $L$ and the angle $\theta$ as marked.  Hence
the longer leg as shown (the adjacent side) has length $L\cos(\theta)$, while
the other one has length $L\sin(\theta)$.  The triangle
at the bottom is congruent to this one, so it has the same side lengths.  In the same way, 
the smaller triangles on the sides have hypotenuse $H$ and legs 
$H\cos(\theta)$ and $H\sin(\theta)$.  Now the whole red rectangle is dissected
into four triangles and the blue rectangle, so its total area is
$$A = HL + 2\cdot \frac{1}{2}L^2 \cos(\theta)\sin(\theta) + 2\cdot \frac{1}{2} H^2 \cos(\theta)\sin(\theta).$$
This simplifies to 
$$A = HL + (L^2 + H^2)\cos(\theta)\sin(\theta).$$
To find the maximum area, we differentiate with respect to $\theta$, set $= 0$ and solve:
$$\frac{dA}{d\theta} = (L^2+H^2)(\cos^2(\theta) - \sin^2(\theta)) = 0$$
when $\cos^2(\theta) = \sin^2(\theta)$, or $\cos(\theta) = \pm \sin(\theta)$.  
It is clear from the diagram that 
only $\theta \in [0,\pi/2]$ are reasonable solutions, so there is just one relevant
critical point, namely the solution of $\cos(\theta) = \sin(\theta)$ in that interval:  
$\theta = \frac{\pi}{4}$.  This gives a maximum of the area $A$ by the First Derivative
Test:  If $0 < \theta < \pi/4$, then $\cos(\theta) > \sin(\theta)$, so $\frac{dA}{d\theta} >0$,
while if $\pi/4 < \theta < \pi/2$, $\cos(\theta) < \sin(\theta)$ and $\frac{dA}{d\theta} < 0$.
(You could also use the Second Derivative Test or the $A$ function values at $\theta = 0,
\pi/4, \pi/2$ to get the same conclusion.)  Hence the maximum area is 
$$A = HL + (L^2 + H^2)\cos(\pi/4)\sin(\pi/4) = HL + \frac{1}{2}(L^2 + H^2).$$


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