OBJECTIVES

When you finish this section, you should be able to:

1. Find Critical Points (p. 880)
2. Use the Second Partial Derivative Test (p. 882)
3. Find the absolute extrema of a function of two variables (p. 883)
4. Solve optimization problems (p. 886)

NEED TO REVIEW?

Local extrema and absolute extrema of a function of a single variable are discussed in Section 4.2, pp. 263-272.

spanDEFINITIONspan Local Maximum, Local Minimum

Let \(z=f(x,y)\) be a function with domain \(D\) and let \((x_{0},y_{0})\) be an interior point of \(D\). Then \(f\) has a local maximum at \( (x_{0},y_{0})\) if \[ \begin{equation*} f(x_{0},y_{0})\geq f(x,y) \end{equation*} \]

for all points \((x,y)\) within some disk centered at \((x_{0},y_{0})\). The number \(f(x_{0},y_{0})\) is called a local maximum value.

The function \(f\) has a local minimum at \((x_{0},y_{0})\) if \[ \begin{equation*} f(x_{0},y_{0})\leq f(x,y) \end{equation*} \]

for all points \((x,y)\) within some disk centered at \((x_{0},y_{0}).\) The number \(f(x_{0},y_{0})\) is called a local minimum value.

Collectively, the local maxima and local minima of \(f\) are called local extrema.

spanDEFINITIONspan Absolute Maximum, Absolute Minimum

Let \(z=f(x,y)\) be a function with domain \(D\). If there is a point \((x_{0},y_{0})\) in \(D\) for which \(f( x_{0},y_{0}) \geq f( x,y)\) for all \(( x,y)\) in \(D\), then \(f\) has an absolute maximum at \(( x_{0},y_{0})\) and \(f(x_{0},y_{0})\) is called the absolute maximum value of \(f\) on \(D\).

If there is a point \(( x_{0},y_{0})\) in \(D\) for which \(f( x_{0},y_{0}) \leq f( x,y)\) for all \(( x,y)\) in \(D\), then \(f\) has an absolute minimum at \(( x_{0},y_{0})\) and \(f( x_{0},y_{0})\) is the absolute minimum value of \(f\) on \(D\).

The absolute maximum and absolute minimum of \(f\) are called the absolute extrema of the function.

THEOREM A Necessary Condition for Local Extrema

Let \(z=f(x,y)\) be a function of two variables that is defined and continuous on an open set containing the point \((x_{0},y_{0})\). Suppose \( f_{x}\) and \(f_{y}\) each exist at \((x_{0},y_{0})\). If \(f\) has a local extremum at \((x_{0},y_{0})\), then \[ {\bbox[#FAF8ED]{\bbox[5px, border:1px solid black, #F9F7ED]{ f_{x}(x_{0},y_{0})=0\qquad \hbox{and} \qquad f_{y}(x_{0},y_{0})=0 }}} \]

spanDEFINITIONspan Critical Point

Let \(z=f(x,y)\) be a function of two variables whose domain is an open set \(D\) containing the point \((x_{0},y_{0})\). The point \((x_{0},y_{0})\) is a critical point of \(f\) if:

• \(f_{x}(x_{0},y_{0})=f_{y}(x_{0},y_{0})=0\).
• either \(f_{x}(x_{0},y_{0})\) or \(f_{y}(x_{0},y_{0})\) does not exist.

THEOREM A Necessary Condition for Local Extrema

Let \(z=f(x,y)\) be a function of two variables whose domain is an open set \(D\). If \(f\) has a local extremum at \((x_{0},y_{0})\), then \((x_{0},y_{0})\) is a critical point of \(f\).

Finding Critical Points

Find the critical points of the function \[ z=f(x,y)=x^{2}+y^{2}-2x+4y \]

Solution  The partial derivatives of \(f\) are \[ f_{x}=2x-2\qquad \hbox{and} \qquad f_{y}=2y+4 \]

Since both partial derivatives exist for all \(x\) and \(y\), the critical points of \(f\) satisfy the system of equations \[ \left\{\begin{array}{l} 2x-2=0\\[3pt] 2y+4=0 \end{array}\right. \]

Solving the system simultaneously, we find that the only critical point is \( (1,-2) \).

NOW WORK

Problem 7.

Finding Critical Points

Consider the hyperbolic paraboloid defined by the equation \( z=f(x,y)=y^{2}-x^{2}\). Show that \((0,0)\) is the only critical point of \(f\), but that \(f\) has neither a local maximum nor a local minimum at \((0,0)\).

Solution  The partial derivatives of \(f\) are \[ f_{x}(x,y)=-2x\qquad \hbox{and} \qquad f_{y}(x,y)=2y \]

At \(( 0,0) \), both partial derivatives equal \(0\), so \((0,0)\) is the only critical point.

See Figure 15. If we consider the values of \(f(x,0)=-x^{2}\), the function attains a maximum value of \(0\) at the origin. However, if we consider the values of \(f(0,y)=y^{2}\), the function attains a minimum value at the origin. In other words, at the critical point \((0,0)\), the function appears to have a maximum when viewed in one direction and to have a minimum when viewed in another direction. As a result, \(z=f(x,y)=y^{2}-x^{2}\) has neither a maximum nor a minimum at (\(0,0\)).

spanDEFINITIONspan Saddle Point

A point \((x_{0},y_{0},f(x_{0},y_{0}))\) is called a saddle point of \(z=f( x,y)\) if both \(f_{x}(x_{0},y_{0})=0\) and \(f_{y}(x_{0},y_{0})=0\), but \(f\) does not have a local extremum at \( (x_{0},y_{0})\).

NOTE

The expression \(AC-B^{2}= f_{xx} f_{yy}-f_{xy}^{2}\) is also called the discriminant or Hessian of \(f\). It is often written as the determinant \[ AC-B^{2} =f_{xx} f_{yy} -f_{xy}^{2} = \left\vert \begin{array}{c@{\quad}c} f_{xx} & f_{xy} \\[3pt] f_{xy} & f_{yy} \end{array} \right\vert \]

THEOREM Second Partial Derivative Test for Local Extrema

Let \(z=f(x,y)\) be a function of two variables for which the first- and second-order partial derivatives are continuous in some disk containing the point \((x_{0},y_{0})\). Suppose \(f_{x}(x_{0},y_{0})=0\) and \(f_{y}(x_{0},y_{0})=0\). Let \[ A=f_{xx}(x_{0},y_{0})\qquad B=f_{xy}(x_{0},y_{0}) \qquad C=f_{yy}(x_{0},y_{0}) \]

1. If \(AC-B^{2}>0\) and \(A>0\), then \(f\) has a local minimum at \( (x_{0},y_{0})\).
2. If \(AC-B^{2}>0\) and \(A<0\), then \(f\) has a local maximum at \( (x_{0},y_{0})\).
3. If \(AC-B^{2}<0\), then \(f\) has a saddle point at \((x_{0},y_{0})\).
4. If \(AC-B^{2}=0\), then the test gives no information.

Using the Second Partial Derivative Test

Find all local maxima, local minima, and saddle points for \[ z=f(x,y)=x^{2}+xy+y^{2}-6x+6 \]

Solution  The critical points of \(f\) are solutions of the system of equations \[ \left\{ \begin{array}{rcll} f_{x}(x,y)&=&2x+y-6&=0 \\[3pt] f_{y}(x,y)&=&x+2y &=0 \end{array} \right. \]

The only solution is \(x=4\) and \(y=-2,\) so \((4,-2)\) is the only critical point.

The second-order partial derivatives of \(f\) are \[ f_{xx}(x,y)=2\qquad f_{xy}(x,y)=f_{yx}( x,y) =1 \qquad f_{yy}(x,y)=2 \]

The function \(f\) meets the requirements of the Second Partial Derivative test at the critical point \((4,-2)\). Now \[ A=f_{xx}(4,-2)=2\qquad B=f_{xy}(4,-2)=1 \qquad C=f_{yy}(4,-2)=2 \] \[ AC-B^{2}=( 2) ( 2) -1^{2}=3>0 \]

Since \(AC-B^{2}>0\) and \(A= f_{xx}(4,-2)=2>0,\) the Second Partial Derivative Test guarantees that \(f\) has a local minimum at \((4,-2)\). The local minimum value is \(z=f(4,-2)=-6\). See Figure 16.

NOW WORK

Problems 13 and 17.

NOTE

To find the critical points of a function \(z=f(x,y)\) we must solve the system of equations \(f_{x} =0\) and \(f_{y} =0\). If the system is nonlinear, no general method of solution is available and a CAS is generally used. If solving by hand, care must be taken because extraneous roots are sometimes introduced.

Using the Second Partial Derivative Test

Find all local maxima, local minima, and saddle points for \[ z=f(x,y)=x^{3}+y^{2}+2xy-4x-3y+5 \]

Solution  The critical points are the solutions of the system of equations \[ \left\{ \begin{array}{rcll} f_{x}(x,y)&=&3x^{2}+2y-4&=0 \\[3pt] f_{y}(x,y)&=&2y+2x-3&=0 \end{array}\right. \]

We solve the system using substitution. Solving for \(y\) in \(f_{y}(x,y)=0\), we obtain \(y=\dfrac{1}{2}(3-2x)\). Then if we substitute for \(y\) in \( f_{x}(x,y)=0\), the result is \[ \begin{eqnarray*} 3x^{2}+2\,\left[ \dfrac{1}{2}(3-2x)\right] -4& =0 \\[4pt] 3x^{2}-2x-1& =0 \\[4pt] (3x+1)(x-1)& =0 \\[4pt] x=-\dfrac{1}{3}\qquad \hbox{or }\qquad x& =1 \end{eqnarray*} \]

When \(x=-\dfrac{1}{3}\), then \(y=\dfrac{1}{2}(3-2x)=\dfrac{1}{2}\left( 3+ \dfrac{2}{3}\right) =\dfrac{11}{6}\).

When \(x=1\), then \(y=\dfrac{1}{2}(3-2x)=\dfrac{1}{2}\).

The critical points are \(\left( -\dfrac{1}{3},\dfrac{11}{6}\right)\) and \( \left( 1,\dfrac{1}{2}\right) \).

The second-order partial derivatives of \(f\) are \[ f_{xx}(x,y)=6x\qquad f_{xy}(x,y)=2\qquad f_{yy}(x,y)=2 \]

The function \(f\) satisfies the requirements of the Second Partial Derivative test at each critical point. At \(\left( -\dfrac{1}{3},\dfrac{11}{6}\right) \), \[ \begin{eqnarray*} &A=f_{xx}\left( -\dfrac{1}{3},\dfrac{11}{6}\right) =-2\qquad B=f_{xy}\left( -\dfrac{1}{3},\dfrac{11}{6}\right) =2\qquad C=f_{yy}\left( -\dfrac{1}{3},\dfrac{11}{6}\right) =2&\\[4pt] &AC-B^{2}=-4-4=-8<0& \end{eqnarray*} \]

So, \(\left( -\dfrac{1}{3},\dfrac{11}{6},\dfrac{317}{108}\right)\) is a saddle point of \(f\). At \(\left( 1,\dfrac{1}{2}\right) \), \[ A=f_{xx}\left( 1,\dfrac{1}{2}\right) =6\qquad B=f_{xy}\left( 1,\dfrac{1}{2}\right) =2\qquad C=f_{yy}\left( 1,\dfrac{1}{2}\right)=2 \]

Since \(AC-B^{2}=8>0\), \(f\) has a local minimum at \(\left( 1,\dfrac{1}{2}\right) \), where \(f\left( 1,\dfrac{1}{2}\right) =\dfrac{7}{4}\)

NOW WORK

Problems 19 and 23.

NEED TO REVIEW?

Open and closed sets are discussed in Section 12.2, p. 825.

THEOREM Extreme Value Theorem for Functions of Two Variables

Let \(z=f(x,y)\) be a function of two variables. If \(f\) is continuous on a closed, bounded set \(D\), then \(f\) has an absolute maximum and an absolute minimum on \(D\).

THEOREM Test for Absolute Maximum and Absolute Minimum

Let \(z=f(x,y)\) be a function of two variables defined on a closed, bounded set \(D.\) If \(f\) is continuous on \(D\), then the absolute maximum value and the absolute minimum value of \(f\) are, respectively, the largest and smallest values found among the following:

• The values of \(f\) at the critical points of \(f\) in \(D\)
• The values of \(f\) on the boundary of \(D\)

Finding Absolute Extrema

Find the absolute maximum and the absolute minimum of \[ \begin{equation*} z=f(x,y)=2x-2xy+y^{2} \end{equation*} \]

whose domain is the region defined by \(0\leq x\leq 4\) and \(0\leq y\leq 3\).

Solution  The function \(f\) is continuous on its domain, which is a closed, bounded set. Then the Extreme Value Theorem guarantees that \(f\) has an absolute maximum and an absolute minimum on its domain. To find them, we first find the critical points of \(f,\) namely the solutions of the system of equations \[ \left\{\begin{array}{l} f_{x}(x,y)=2-2y=0\\ f_{y}(x,y)=-2x+2y=0 \end{array}\right. \]

Solving the system of equations, we find that the only critical point is \((1,1)\). The value of \(f\) at \((1,1)\) is \[ f(1,1)=2(1) -2(1) (1) +1^{2}=1 \]

The domain of \(f\) is the set \(0\leq x \leq 4\), \(0\leq y\leq 3\). The boundary of the domain of \(f\) consists of the four line segments \(L_1\), \(L_2\), \(L_3\), and \(L_4\) shown in Figure 18. We evaluate \(f\) on each line segment.

• On \(\boldsymbol L_{\bf 1}\): \(x\) is in the interval \(\left[ 0,4\right]\) and \(y=0\). The function \(f(x,y)=f(x,0)=2x\) is increasing on \(0\leq x\leq 4\), so its extreme values occur at the endpoints \(0\) and \(4.\) \[ \hbox{At }x=0, f(0,0)=0\qquad \hbox{and} \qquad \hbox{at}\ x=4, f(4,0)=8 \]

• On \(\boldsymbol L_{\bf 2}\): \(x=4\) and \(y\) is in the interval \([0,3] \). The function \(f(x,y)=f(4,y)= 8-8y+y^{2}\). To find the extreme values of \(f\) on \( [0,3] \), we begin by finding the critical number(s) of the function \(g(y) =8-8y+y^{2}.\) That is, we find where \(g' (y) =0.\) \[ \begin{eqnarray*} g' (y) &=&-8+2y=0 \\[3pt] y &=&4 \end{eqnarray*} \]

  Since \(4\) is not in the interval \([0,3] ,\) the extreme values occur at the endpoints 0 and 3. \[ \hbox{At }y=0\hbox{, }\ f(4,0)=8\qquad \hbox{and} \qquad \hbox{at}y=3,\hbox{ }f(4,3)=-7 \]

• On \(\boldsymbol L_{\bf 3}\): \( x\) is in the interval \([ 0,4] \) and \(y=3\). The function \(f(x,y)=f(x,3)= 2x-6x+9=-4x+9\) is decreasing on \(0\leq x\leq 4\), so its extreme values occur at the endpoints \(0\) and \(4.\) \[ \hbox{At }x=0, f(0,3)=9\qquad \hbox{and} \qquad \hbox{at }x=4, f(4,3)=-7 \]

• On \(\boldsymbol L_{\bf 4}\): \(x=0\) and \(y\) is in the interval \([0,3] .\) The function \(f(x,y)=f(0,y)=y^{2}\) is increasing on \(0\leq y\leq 3\), so its extreme values occur at the endpoints \(0\) and \(3.\) \[ \hbox{At }y=0, f(0,0)=0\qquad\hbox {and}\qquad \hbox{at }y=3,\hbox{ }f(0,3)=9 \]

The values of \(f\) at the critical point \((1,1)\) and at the extreme values on the boundary are

The absolute maximum value of \(f\) is \(f(0,3)=9\); the absolute minimum value is \(f(4,3)=-7\).

NOW WORK

Problem 33.

Finding Absolute Extrema

Find the absolute maximum and the absolute minimum of \[ \begin{equation*} z=f( x,y) =x^{2}+y^{2}-2x+2y-5 \end{equation*} \]

whose domain is the disk \(x^{2}+y^{2}\leq 9\).

Solution  The function \(f\) is continuous on its domain, a closed, bounded set. The Extreme Value Theorem guarantees that \(f\) has an absolute maximum and an absolute minimum on its domain. To find them, we find the critical points of \(f,\) namely the solutions of the system of equations \[ \left\{\begin{array}{l} f_{x}( x,y) =2x-2=0\\[3pt] f_{y}( x,y) =2y+2=0 \end{array}\right. \]

The only solution is \(x=1,\) \(y=-1,\) which is an interior point of the domain of \(f.\) The value of \(f\) at the critical point is \(f(1,-1) =-7\).

The boundary of the domain of \(f\) is the circle \(x^{2}+y^{2}=9.\) We express the boundary using the parametric equations \(x=x(t) =3\,\cos t\), \(y=y(t) =3\,\sin t,\) \(0\leq t\leq 2\pi .\) Next we evaluate \(f\) on its boundary. \[ \begin{eqnarray*} z=f( x,y) &=&f( 3\,\cos t,3\,\sin t)\\[4pt] &=&9\,\cos ^{2}t+9\,\sin^{2}t-6\,\cos t+6\,\sin t-5\\[4pt] &=&4-6\,\cos t+6\,\sin t \end{eqnarray*} \]

Now we find where \(\dfrac{dz}{dt}=0\). \[ \begin{eqnarray*} \dfrac{dz}{dt}&= & 6\,\sin t+6\,\cos t=0 \\[4pt] \tan t&=&-1 \\[4pt] t&=&\dfrac{3\pi }{4}\qquad\hbox{or}\qquad t=\dfrac{7\pi }{4} \end{eqnarray*} \]

At \(t=\dfrac{3\pi }{4},\) the value of \(f\) is \[ \begin{eqnarray*} f\left( 3\,\cos \dfrac{3\pi }{4},3\,\sin \dfrac{3\pi }{4}\right) &=&4-6\,\cos \dfrac{ 3\pi }{4}+6\,\sin \dfrac{3\pi }{4}=4-6\left( -\dfrac{\sqrt{2}}{2}\right) +6\left( \dfrac{\sqrt{2}}{2}\right)\\[4pt] &=&4+6\sqrt{2} \end{eqnarray*} \]

At \(t=\dfrac{7\pi }{4},\) the value of \(f\) is \[ \begin{eqnarray*} f\left( 3\,\cos \dfrac{7\pi }{4},3\,\sin \dfrac{7\pi }{4}\right) &=&4-6\,\cos \dfrac{ 7\pi }{4}+6\,\sin \dfrac{7\pi }{4}=4-6\left( \dfrac{\sqrt{2}}{2}\right) +6\left( -\dfrac{\sqrt{2}}{2}\right)\\[4pt] &=&4-6\sqrt{2} \end{eqnarray*} \]

The absolute maximum value of \(f\) is \(4+6\sqrt{2}\approx 12. 485;\) the absolute minimum value of \(f\) is \(-7.\)

NOW WORK

Problem 35.

Solving an Optimization Problem: Minimizing Resources Used

A manufacturer wants to make an open rectangular box of volume \(V=500\) cubic centimeters (\(\rm{cm}^3\)) using the least possible amount of material. Find the dimensions of the box.

Solution  Let \(x\) and \(y\) be the dimensions of the base of the box, and let \(z\) be the height of the box. Then \[ V=500=xyz\,\rm{cm}^{3}\qquad x>0\quad y>0\quad z>0 \]

The manufacturer wants to minimize the amount of material used, which equals the surface area \(S\) of the box. Since the box is open, \[ S=xy+2xz+2yz \]

If we solve the equation \(500=xyz\) for \(z\) and substitute \(z=\dfrac{500}{xy}\) into the formula for the surface area \(S\) of the box, we can express \(S\) as a function of two variables. \[ S=S( x,y) =xy+2x\!\left( \dfrac{500}{xy}\right) +2y\left( \dfrac{500 }{xy}\right) =xy+\frac{1000}{y}+\frac{1000}{x} \]

This is the function to be minimized. The partial derivatives of \(S\) are \[ S_{x}=y-\dfrac{1000}{x^{2}}\qquad \hbox{and} \qquad S_{y}=x-\dfrac{1000}{y^{2}} \]

Since \(x>0\) and \(y>0,\) the critical points satisfy the system of equations \[ \left\{\begin{array}{l} y-\dfrac{1000}{x^{2}}=0\\[3pt] x-\dfrac{1000}{y^{2}}=0 \end{array}\right. \]

Using substitution, we find \[ x=y=1000^{1/3}=10 \]

The second-order partial derivatives of \(S\) are \[ \begin{eqnarray*} S_{xx}&=&\dfrac{\partial }{\partial x}\left( y-\dfrac{1000}{x^{2}}\right) =\dfrac{2000}{x^{3}}\qquad S_{xy}= \dfrac{\partial }{\partial x}\left(x-\dfrac{1000}{y^{2}}\right) =1\\[4pt] S_{yy}&=&\dfrac{\partial }{\partial y}\left( x-\dfrac{1000}{y^{2}}\right) =\dfrac{2000}{y^{3}} \end{eqnarray*} \]

At the critical point \(( 10,10) \), \[ \begin{eqnarray*}\hspace{-1pc} &\hspace{-1.8pc}A=S_{xx}(10,10) =\dfrac{2000}{10^{3}}=2\qquad B=S_{xy}(10,10) =1\qquad C=S_{yy}(10,10) =\dfrac{2000}{10^{3}}=2 &\hspace{1.8pc}\\[4pt] &AC-B^{2}=3>0& \end{eqnarray*} \]

From the Second Partial Derivative Test, \(S\) has a local minimum at \((10,10)\).

But, we want to know where \(S\) attains its absolute minimum. Since the physical properties of the problem require that the absolute minimum exists, the local minimum is also the absolute minimum. Consequently, the dimensions (in centimeters) of the open box of volume \(V=500\,\rm{cm}^{3}\) that uses the least amount of material are \[ x=10\rm{cm}\qquad y=10\rm{cm}\qquad z=\dfrac{500}{100}=5\rm{cm} \]

NOW WORK

Problem 43.

Solving an Optimization Problem: Maximizing Profit

The demand functions for two products are \[ p=12-2x\qquad \hbox{and}\qquad q=20-y \]

where \(p\) and \(q\) are the respective prices (in thousands of dollars) of each product, and \(x\) and \(y\) are the respective amounts (in thousands of units) of each sold. Suppose the joint cost function is \[ C( x,y) =x^{2}+2xy+2y^{2} \]

1. Find the revenue function and the profit function.
2. Determine the prices and amounts that will maximize profit.
3. What is the maximum profit?

Solution  (a) Revenue is the amount of money brought in. That is, revenue is the product of price and quantity sold. The revenue function \(R\) is the sum of the revenues from each product. \[ R( x,y) =xp+yq=x( 12-2x) +y( 20-y) \]

Profit is the difference between revenue and cost. The profit function \(P\) is \[ \begin{eqnarray*} P( x,y) &=&R( x,y) -C( x,y) =[ x(12-2x) +y( 20-y) ] -[ x^{2}+2xy+2y^{2}] \\[3pt] &=&12x-2x^{2}+20y-y^{2}-x^{2}-2xy-2y^{2}\\[3pt] &=&-3x^{2}-3y^{2}-2xy+12x+20y \end{eqnarray*} \]

(b) The partial derivatives of \(P\) are \[ \begin{eqnarray*} P_{x}( x,y) &=&\dfrac{\partial }{\partial x}(-3x^{2}-3y^{2}-2xy+12x+20y) =-6x-2y+12 \\[4pt] P_{y}( x,y) &=&\dfrac{\partial }{\partial y}(-3x^{2}-3y^{2}-2xy+12x+20y) =-6y-2x+20 \end{eqnarray*} \]

We find the critical points by solving the system of equations: \[ \left\{ \begin{array}{l} -6x-2y+12=0 \\[3pt] -2x-6y+20=0 \end{array} \right. \]

We solve the first equation for \(y\) and substitute the result into the second equation. Since \(y=6-3x,\) then \[ \begin{eqnarray*} -2x-6( 6-3x) +20 &=&0 \\ x &=&1 \end{eqnarray*} \]

and \(y=6-3(1) =3\), so \((1,3) \) is the only critical point.

The second-order partial derivatives are \[ A=P_{xx}( x,y) =-6\qquad B=P_{xy}( x,y) =-2\qquad C=P_{yy}( x,y) =-6 \]

At the critical point \((1,3) ,\) \[ \begin{array}{@{\hspace*{-3.4pc}}l} P_{xx}(1,3) =-6<0 \qquad P_{xy}(1,3) =-2 \qquad P_{yy}(1,3) =-6 \qquad AC-B^{2}=36-4=32>0 \end{array} \]

From the Second Partial Derivative Test, \(P\) has a local maximum at \((1,3)\).

The domains of the demand functions \(p=12-2x\) and \(q=20-y\) are \(0\leq x\leq 6\) and \(0\leq y\leq 20\), respectively. Since the domain of the profit function \(P\) is \(0\leq x\leq 6,\) \(0\leq y\leq 20,\) the domain is closed and bounded, so a maximum profit exists. From the test for absolute maximum and absolute minimum, the maximum profit is found at a critical point or on the boundary of the domain.

Figure 20 shows the boundary of the profit function \(P\) and Table 1 gives the maximum value of \(P\) on its boundary and at the critical point (\(1,3\)).

We conclude that by selling \(1000\) units of product \(x\) for the price \( p=12-2(1) =10\) thousand dollars and \(3000\) units of product \(y\) for the price \(q=20-3=17\) thousand dollars, the profit is maximized.

(c) The maximum profit is \(P(1,3) =$ 36,000.\)

NOW WORK

Problem 55.