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wlad13 [49]
3 years ago
15

A RSS Clothing manufacturer makes two types of jogging pants, design A and design B. The design A jogging pants sells to the ret

ail stores for 2,500each and the design B jogging pants for 2,100. The cost for manufacturing each design A is 1,750 and the cost of each design B is 1,200. If the manufacturer makes no more than 120 jogging a week and budget no more than 150,000 per week, how many of each type should be made to maximize profit?

Mathematics
1 answer:
nikdorinn [45]3 years ago
8 0

Answer:

  120 of design B

Step-by-step explanation:

The profit per dollar cost for design A is (2500/1750 -1) ≈ 0.43. For design B, it is (2100/1200 -1) ≈ 0.75. Design B is more profitable and less expensive to make, so its production should be maximized.

For maximum profit, 120 jogging pants of design B should be made each week.

_____

The budget of 150,000 would allow 150,000/1,200 = 125 pants of design B to be made. The production limit of 120 pants does not use the entire budget.

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The graph of g(x) = -x^2 is shifted 3 units left and 1 unit up. If this new graph is f(x), then what is the value of f(-1.5)
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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}\\\\
--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see.

to the left by 3 units, C = +3
up by 1  unit, D = 1

\bf g(x)=-x^2\\\\\\ g(x)=-1(1x+\stackrel{C}{0})^2+\stackrel{D}{0}\implies g(x)=-1(1x+3)^2+1
\\\\\\
g(x)=-(x+3)^2+1\impliedby f(x)\qquad \qquad f(-1.5)=-[(-1.5)+3]^2+1

and surely you know how much that is.
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