2.9 Answers to the Additional Problems

  1. \(\partial z/\partial x=(0.00187y^{0.7})=\)constant for given \(y\).
    Exposed profile should be a straight line.
    \(\partial z/\partial y=(0.00187)(x+1)(0.7)y^{-0.3}\)
    Exposed profile curves upward with decreasing slope.

  2. \(\partial z/\partial x=2(P1)x\), \(\partial z/\partial y=2(P2)y\)
    \(\partial^2 z/\partial x^2=2(P1)\), \(\partial^2z/\partial x\partial y=0\), \(\partial^2z/\partial y^2=2(P2)\)
    \(\partial z/\partial x=0\) if \(x=0\)
    \(\partial z/\partial y=0\) if \(y=0\)
    Thus \((x,y)=(0,0)\) is the only critical point.
    To classify the critical point, evaluate: \[\frac{\partial^2z}{\partial x^2}\frac{\partial^2z}{\partial y^2}-\Big(\frac{\partial^2 z}{\partial x\partial y}\Big)^2=4(P1)(P2)\] Pick \(P1=.6\), \(P2=1.0\)
    Then \(4(P1)(P2)>0\), \(\frac{\partial^2z}{\partial x^2}=2(P1)>0\) and the point is a relative minimum.

  3. We use the Pythagorean theorem to obtain

\[(2r)^2=W^2+D^2\] Thus \[\begin{align*} D^2&=4r^2-W^2 \\ S&=(0.1)4r^2W-(0.1)W^3 \\ \frac{dS}{dW}&=(0.4)r^2-(0.3)W^2 \\ 0&=(0.4)r^2-(0.3)W^2 \end{align*}\]

The critical point is then at \(W=\sqrt{4r^2/3}=2r/\sqrt{3}\) We have \[\frac{d^2S}{dW^2}=-0.6W<0\] so that when \(W=2r/\sqrt{3}\), \(S\) is indeed at a maximum.
The depth is then \[\begin{align*} D&=\sqrt{4r^2-W^2} \\ &=\sqrt{4r^2-4r^2/3} \\ &=2\sqrt{2/3}r \end{align*}\]

4.a. Even assuming \(r = r(t)\), N cannot exceed \(K\) if \(N(0) < K\). We see this by writing \(N\) as a fraction. \[N(t)=\frac{K}{1+be^{-r(t)t}}\] b. Treat \(r(t)\) in the difference equation model, with \[\begin{align*} r(0)&=r_0 \\ r(1)&=r_0/2 \\ r(2)&=r_0/3 \\ r(3)&=r(2) \end{align*}\] or some such scheme to decrease r as time increases.

  1. Evaluate the derivative. \[Y'=(P1)X-(P2)X^2\] The “rate of increase” of \(Y\) is greatest when \(Y'\) is maximal. Find, the max(\(Y'\)) by differentiating \(Y'\) and setting \(Y'' = 0\). \[\begin{align*} Y''&=\frac{d(Y')}{dx}=(P1)-(2)(P2)X \\ 0&=(P1)-(2)(P2)X \\ X&=(P1)/(2(P2)) \end{align*}\] The value for \(Y\) is then \[Y=\Big[\frac{P1}{2(P2)}\Big]^2\Big[\frac{P1}{2}-\frac{(P1)(P2)}{6(P2)}\Big]=\frac{(P1)^3}{12(P2)^2}\] This point is an inflection point on the graph \(Y\) versus \(X\).