3.2: Graphical Solutions
- Page ID
- 1813
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Before giving a more general algorithm for handling this problem and problems like it, we note that when the number of variables is small (preferably 2), a graphical technique can be used.
Inequalities, such as the four given in Pablo's problem, are often called \(\textit {constraints}\), and values of the variables that satisfy these constraints comprise the so-called \(\textit {feasible region}\). Since there are only two variables, this is easy to plot:
Example \(\PageIndex{1}\): Constraints and feasible region
Pablo's constraints are
\begin{eqnarray*}
&x\geq 5&\\
&y\geq 7&\\[2mm]
&15\leq x+y\leq25\, .&
\end{eqnarray*}
Plotted in the \((x,y)\) plane, this gives:
You might be able to see the solution to Pablo's problem already. Oranges are very sugary, so they should be kept low, thus \(y=7\). Also, the less fruit the better, so the answer had better lie on the line \(x+y=15\). Hence, the answer must be at the vertex \((8,7)\). Actually this is a general feature of linear programming problems, the optimal answer must lie at a vertex of the feasible region. Rather than prove this, lets look at a plot of the linear function \(s(x,y)=5x+10y\).
Example \(\PageIndex{2}\): The sugar function
Plotting the sugar function requires three dimensions:
The plot of a linear function of two variables is a plane through the origin. Restricting the variables to the feasible region gives some lamina in 3-space. Since the function we want to optimize is linear (and assumedly non-zero), if we pick a point in the middle of this lamina, we can always increase/decrease the function by moving out to an edge and, in turn, along that edge to a corner. Applying this to the above picture, we see that Pablo's best option is 110 grams of sugar a week, in the form of 8 apples and 7 oranges.
It is worthwhile to contrast the optimization problem for a linear function with the non-linear case you may have seen in calculus courses:
Here we have plotted the curve \(f(x)=d\) in the case where the function \(f\) is linear and non-linear. To optimize \(f\) in the interval \([a,b]\), for the linear case we just need to compute and compare the values \(f(a)\) and \(f(b)\). In contrast, for non-linear functions it is necessary to also compute the derivative \(\frac{\mathrm{d} f}{\mathrm{d} x}\) to study whether there are extrema inside the interval.
Contributor
David Cherney, Tom Denton, and Andrew Waldron (UC Davis)