Two artists make winter yard ornaments.They get $83 for each wooden snowman they make and $74 for each wooden Santa Claus.On average, Nina must work 8 hours and Roberta 6 hours to make a
Snowman.Nina must work 4 hours and Roberta 8 hours to make a Santa Claus.If neither wishes to
Work more than 40 hours per week, how many of each ornament should they make each week to
Maximize their income? The information is summarized in the table below.

A) 4 snowmen and 3 ornaments of Santa Claus: $\$ 34$
B) $\quad 3$ snowmen and 4 ornaments of Santa Claus: $\$ 34$
C) 4 snowmen and 2 ornaments of Santa Claus: $\$ 480$
D) $\quad 2$ snowmen and 5 ornaments of Santa Claus: $\$ 46$
E) $\quad 5$ snowmen and no ornaments of Santa Claus: $\$ 415$

Question

Two artists make winter yard ornaments.They get $83 for each wooden snowman they make and $74 for each wooden Santa Claus.On average, Nina must work 8 hours and Roberta 6 hours to make a
Snowman.Nina must work 4 hours and Roberta 8 hours to make a Santa Claus.If neither wishes to
Work more than 40 hours per week, how many of each ornament should they make each week to
Maximize their income? The information is summarized in the table below.

A) 4 snowmen and 3 ornaments of Santa Claus: $\$ 34$ 
B) $\quad 3$ snowmen and 4 ornaments of Santa Claus: $\$ 34$ 
C) 4 snowmen and 2 ornaments of Santa Claus: $\$ 480$ 
D) $\quad 2$ snowmen and 5 ornaments of Santa Claus: $\$ 46$ 
E) $\quad 5$ snowmen and no ornaments of Santa Claus: $\$ 415$

Quizplus · Accepted Answer

Let the number of snowmen made by Nina and Roberta be $x$ and the number of Santa Claus made by Nina and Roberta be $y$. Then the objective is to maximize the income $I=83x+74y$, subject to the following constraints:
\begin{align*}
8x+6y&\leq 40 \
4x+8y&\leq 40 \
x,y&\geq 0
\end{align*}
The first two constraints come from the fact that Nina and Roberta cannot work more than 40 hours per week. 
Solving the system of inequalities, we get the feasible region shown below:
[asy]
import graph; size(7cm); real lsf=0.5;
draw((0,0)--(0,8.33),Arrow); draw((0,0)--(9.17,0),Arrow); label("$x$",(9.17,0),SE*lsf); label("$y$",(0,8.33),NE*lsf);
dot((0,6),linewidth(4pt));label("$(0,6)$",(0,6),NE*lsf);dot((5,0),linewidth(4pt));label("$(5,0)$",(5,0),NE*lsf); dot((3,3)); dot((4,2)); dot((4,3)); dot((4,4)); dot((3,5)); fill((3,3)--(4,2)--(5,0)--(0,0)--(0,6)--cycle,gray(.7));draw((0,6)--(5,0),linetype("2 2"));
[/asy]
We want to find the point $(x,y)$ in the feasible region that maximizes $I=83x+74y$. Note that $I$ is a linear function, so it is maximized at one of the vertices of the feasible region. We evaluate $I$ at each vertex as shown below:

\begin{tabular}{c|c|c|c|c|c}
$(x,y)$ & $(0,6)$ & $(5,0)$ & $(3,3)$ & $(4,2)$ & $(4,4)$ \ \hline
$I=83x+74y$ & $498$ & $415$ & $477$ & $495$ & $581$
\end{tabular}

Thus, the maximum income is achieved when making 4 snowmen and 2 ornaments of Santa Claus, which gives $I=83(4)+74(2)=\boxed{480}$.

Two Artists Make Winter Yard Ornaments $34\$ 34$34 B) 3\quad 33 Snowmen and 4 Ornaments of Santa Claus

Two Artists Make Winter Yard Ornaments $\$ 34$ B) $\quad 3$ Snowmen and 4 Ornaments of Santa Claus