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    <name>
      <text>18.3.2.3 Location of stationary points (copy)</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<p>Find a second stationary point of \(f(x,y)=8x^2+6y^2-2y^3+5\).</p>

<h4 class="HELM_solution">Solution</h4>
\(f_x=16x\) and \(f_y\equiv 6y(2-y)\).
<p>From this we note that \(f_x=0\) when \(x=0\), and \(f_y=0\) when \(y=0\).</p>
<p>So \(x=0,\ y=0\)</p>
<p> i.e. \((0,0)\) is a second stationary point of the function.</p>
<hr>

<p>It is important when solving the simultaneous equations \(f_x=0\) and \(f_y=0\) to find stationary points not to miss any solutions. A useful tip is to factorise the left-hand sides and consider systematically all the possibilities.</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Locate the stationary points of \(f(x,y)=x^4+y^4-36xy\)</p>

<h4 class="HELM_solution">Solution</h4>
<p>First we write down the partial derivatives of \(f(x,y)\)</p>
<p>\(\dfrac{\partial f}{\partial x}=4x^3-36y=4(x^3-9y)\)</p>
<p>\(\dfrac{\partial f}{\partial y}=4y^3-36x=4(y^3-9x)\)</p>
<p>Now we solve the equations \(\dfrac{\partial f}{\partial x}=0\) and \(\dfrac{\partial f}{\partial y}=0\):
</p>
<p>\(x^3-9y = 0 \quad \) (i)</p>
<p>\(y^3-9x = 0 \quad \) (ii) </p>
<p>From (ii) we obtain</p>
<p>\(x = \dfrac{y^3}{9} \quad\) (iii) </p>
<p>Now substitute from (iii) into (i) </p>

\(\dfrac{y^9}{9^3}-9y = 0\\ \) \(\Rightarrow y^9-9^4y = 0\\ \)
<p>\(\Rightarrow y(y^8-3^8) = 0 \) (removing the common factor) \( \\ \)</p>
<p>\(\Rightarrow y(y^4-3^4)(y^4+3^4) = 0 \) (using the difference of two squares)\( \\ \)</p>
<p>We therefore obtain as the only solutions:</p>
<p>\(y=0\) or \(y^4-3^4 = 0\) (since \(y^4+3^4\) is never zero) \( \\ \)</p>
<p>The last equation implies:</p>
<p>\((y^2-9)(y^2+9) = 0\) (using the difference of two squares) \( \\ \)</p>
<p>therefore \(y^2=9\) and \(y =\pm\ 3\).</p>
<p></p>
<p>Now, using (iii): when \(y=0,\ x=0\), when \(y=3,\ x=3\), and when \(y=-3,\ x=-3\).
</p>
<p>The stationary points are \((0,0), (-3,-3)\) and \((3,3)\).</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Locate the stationary points of \(f(x,y)=x^4+y^4-36xy\)</p>

<h4 class="HELM_solution">Solution</h4>
\(\dfrac{\partial f}{\partial x}=3x^2-3,\\ \dfrac{\partial f}{\partial y}=2y-6\)
<p>Now solve simultaneously the equations \(\dfrac{\partial f}{\partial x}=0\) and \(\dfrac{\partial f}{\partial y}=0\):</p>
\(3x^2-3=0\) and \(2y-6=0\).
<p></p>
<p>Hence \(x^2=1\) and \(y=3\), giving stationary points at \((1,3)\) and \((-1,3)\).
</p>
<hr>



<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p><strong>The Partial Derivative of  \(\pmb f\) with respect to \(\pmb x\)</strong></p>
    <p> For a function of two variables \(z=f(x,y)\) the partial derivative of \(f\) with respect to \(x\) is denoted by \(\dfrac{\partial f}{\partial x}\) and is obtained by differentiating \(f(x,y)\) with respect to \(x\) in the usual way but treating the
        \(y\)-variable as if it were a constant.</p>
    <p>Alternative notations for \(\dfrac{\partial f}{\partial x}\) are \(f_x(x,y)\) or \(f_x\) or \(\dfrac{\partial z}{\partial x}\)
    </p>
</div>

<hr>
<h4 class="HELM_example">Example</h4>
<ol class="HELM_parts_inline">
    <p>Find \(\dfrac{\partial f}{\partial x}\) for</p>
    <li> \(f(x,y)=x^3+x+y^2+y\), </li>
    <li> \(f(x,y)=x^2y+xy^3\).</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>\(\dfrac{\partial f}{\partial x}=3x^2+1+0+0=3x^2+1\)</li>
    <li>\(\dfrac{\partial f}{\partial x}=2x\times y+1\times y^3=2xy+y^3\)</li>
</ol>]]></text>
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