Need Help With a Worded Math Problem.

[Samvineet]

Hi Guys, I really appriciate your time and effort in solving this problem. Thanks in advance.

Products 1 and 2 are produced by use of three machines: A, B, and C.

Each unit of product 1 requires 1 hour on machine A and 2 hours on machine B.

Each unit of product 2 requires 1 hour of time on each machine.

The time available on these machines is limited to 400 hours per month on machine A, 600 hours per month on machine B, and 300 hours per month on machine C.

Each unit of product 1 can be sold to yield a profit of 50 cents and each unit of product B can be sold to yield a profit of 80 cents.
Derive a calculation that can be used to help determine how many of each should be produced each month to maximize profit.

 

[Deleted member 4993]

Hi Guys, I really appriciate your time and effort in solving this problem. Thanks in advance.

Products 1 and 2 are produced by use of three machines: A, B, and C.

Each unit of product 1 requires 1 hour on machine A and 2 hours on machine B.

Each unit of product 2 requires 1 hour of time on each machine.

The time available on these machines is limited to 400 hours per month on machine A, 600 hours per month on machine B, and 300 hours per month on machine C.

Each unit of product 1 can be sold to yield a profit of 50 cents and each unit of product B can be sold to yield a profit of 80 cents.
Derive a calculation that can be used to help determine how many of each should be produced each month to maximize profit.

To start of name your variables. What are the input-variables? What are the output-variables?

Please show us your work, indicating excatly where you are stuck - so that we know where to begin to help you.

 

[Samvineet]

Requirements The Time Available
Machine Product 1 Product 2 Per hour per month
A 1 1 400
B 2 1 600
C 300

Let X1= number of product of type 1

X2= number of product of type 2

Since the product are positive quantity . therefore X1 X2 ? 0

For Machine A, X1 + X2 ? 400
For Machine B 2X1 + X2 ? 600


so... what's next??

 

[Deleted member 4993]

Requirements The Time Available
Machine Product 1 Product 2 Per hour per month
A 1 1 400
B 2 1 600
C 300

Let X[sub:zij4uarh]1[/sub:zij4uarh]= number of product of type 1

X[sub:zij4uarh]2[/sub:zij4uarh]= number of product of type 2

Since the product are positive quantity . therefore X1 X2 ? 0

For Machine A, X[sub:zij4uarh]1[/sub:zij4uarh] + X[sub:zij4uarh]2[/sub:zij4uarh] ? 400
For Machine B 2X[sub:zij4uarh]1[/sub:zij4uarh] + X[sub:zij4uarh]2[/sub:zij4uarh] ? 600
For machine C X[sub:zij4uarh]2[/sub:zij4uarh] ? 300

so... what's next??

What method have you been taught to solve these inequalities?

With two variables, easiest method is - graphical.

Do you know this method?

 

[soroban]

Hello, Samvineet!

Your work is slightly flawed.
My table is slightly different.


\(\displaystyle \text{Let: }\;\begin{array}{ccc}x &=& \text{units of Product 1} \\ y &=& \text{units of Product 2} \end{array}\)


. . \(\displaystyle \begin{array}{c||c|c|c|} & A & B & C \\ \hline\hline \text{Product 1} & x & 2x & \\ \hline \text{Product 2} & y & y & y \\ \hline \text{Total} & 400 & 600 & 300 \\ \hline\end{array}\)


\(\displaystyle \text{We have: }\;\begin{Bmatrix}x \;\geq\;0 & [1] \\ y \;\geq\; 0 & [2] \\ x + y \;\leq\; 400 & [3] \\ 2x + y \;\leq\; 600 & [4] \\ y \:\leq\: 300 & [5]\end{Bmatrix}\)


\(\displaystyle \text{[1] and [2] places us in Quadrant 1.}\)

\(\displaystyle \text{To graph the region of [3], graph the }line,\;x + y \:=\:400\)
. . \(\displaystyle \text{It has interecepts: }\400,0),\;(0,400).\)
\(\displaystyle \text{Draw the line and shade the region below it.}\)

\(\displaystyle \text{To graph the region of [4], graph the }line,\:2x+y \:=\:600\)
. . \(\displaystyle \text{It has intercepts" }\300,),\;(0,600)\)
\(\displaystyle \text{Draw the line and shade the region below it.}\)

\(\displaystyle \text{To graph the region of [5], graph the }line,\:y \:=\:300\)
. . \(\displaystyle \text{This is a horizontal line.}\)
\(\displaystyle \text{Draw the line and shade the region below it.}\)


\(\displaystyle \text{The graph looks something like this:}\)

        |
        |     (100,300)
(0,300) o * * o
        |:::::::*
        |:::::::::*
        |:::::::::::o (200,200)
        |::::::::::::*
        |:::::::::::::*
        |::::::::::::::*
  - - - o - - - - - - - o - - -
        |            (300.0)
        |

\(\displaystyle \text{where }you\text{ must find the vertices of the region.}\)


\(\displaystyle \text{Then substitute the vertices into the Profit function: }\;P \:=\:0.50x + 0.80y\)
. . \(\displaystyle \text{and determine which vertex produces maximum profit.}\)

 

[Samvineet]

Requirements The Time Available
Machine Product 1 Product 2 Per hour per month
A 1 1 400
B 2 1 600
C 300

Let X[sub:2sbppdhj]1[/sub:2sbppdhj]= number of product of type 1

X[sub:2sbppdhj]2[/sub:2sbppdhj]= number of product of type 2

Since the product are positive quantity . therefore X1 X2 ? 0

For Machine A, X[sub:2sbppdhj]1[/sub:2sbppdhj] + X[sub:2sbppdhj]2[/sub:2sbppdhj] ? 400
For Machine B 2X[sub:2sbppdhj]1[/sub:2sbppdhj] + X[sub:2sbppdhj]2[/sub:2sbppdhj] ? 600
For machine C X[sub:2sbppdhj]2[/sub:2sbppdhj] ? 300

so... what's next??

What method have you been taught to solve these inequalities?

With two variables, easiest method is - graphical.

Do you know this method?

No, unfortunately I don't know this method. Could your explain it to me please?

 

[Samvineet]

Hello, Samvineet!

Your work is slightly flawed.
My table is slightly different.


\(\displaystyle \text{Let: }\;\begin{array}{ccc}x &=& \text{units of Product 1} \\ y &=& \text{units of Product 2} \end{array}\)


. . \(\displaystyle \begin{array}{c||c|c|c|} & A & B & C \\ \hline\hline \text{Product 1} & x & 2x & \\ \hline \text{Product 2} & y & y & y \\ \hline \text{Total} & 400 & 600 & 300 \\ \hline\end{array}\)


\(\displaystyle \text{We have: }\;\begin{Bmatrix}x \;\geq\;0 & [1] \\ y \;\geq\; 0 & [2] \\ x + y \;\leq\; 400 & [3] \\ 2x + y \;\leq\; 600 & [4] \\ y \:\leq\: 300 & [5]\end{Bmatrix}\)


\(\displaystyle \text{[1] and [2] places us in Quadrant 1.}\)

\(\displaystyle \text{To graph the region of [3], graph the }line,\;x + y \:=\:400\)
. . \(\displaystyle \text{It has interecepts: }\400,0),\;(0,400).\)
\(\displaystyle \text{Draw the line and shade the region below it.}\)

\(\displaystyle \text{To graph the region of [4], graph the }line,\:2x+y \:=\:600\)
. . \(\displaystyle \text{It has intercepts" }\300,),\;(0,600)\)
\(\displaystyle \text{Draw the line and shade the region below it.}\)

\(\displaystyle \text{To graph the region of [5], graph the }line,\:y \:=\:300\)
. . \(\displaystyle \text{This is a horizontal line.}\)
\(\displaystyle \text{Draw the line and shade the region below it.}\)


\(\displaystyle \text{The graph looks something like this:}\)

        |
        |     (100,300)
(0,300) o * * o
        |:::::::*
        |:::::::::*
        |:::::::::::o (200,200)
        |::::::::::::*
        |:::::::::::::*
        |::::::::::::::*
  - - - o - - - - - - - o - - -
        |            (300.0)
        |

\(\displaystyle \text{where }you\text{ must find the vertices of the region.}\)


\(\displaystyle \text{Then substitute the vertices into the Profit function: }\;P \:=\:0.50x + 0.80y\)
. . \(\displaystyle \text{and determine which vertex produces maximum profit.}\)

Thank you so much for all the help Soroban. When you say " must find the vertices of the region" do you mean something like this -

 

[michaeldoe65]

the giant ocean tank at new england is a cylinder that has a radius of 18ft and a volume of 10178.76? what is the height of the tank to the nearest foot. use 3.14 for pi

 

[Deleted member 4993]

Hello, Samvineet!

\(\displaystyle \text{The graph looks something like this:}\)

        |
        |     (100,300)
(0,300) o * * o
        |:::::::*
        |:::::::::*
        |:::::::::::o (200,200)
        |::::::::::::*
        |:::::::::::::*
        |::::::::::::::*
  - - - o - - - - - - - o - - -
        |            (300.0)
        |

\(\displaystyle \text{where }you\text{ must find the vertices of the region.}\)


\(\displaystyle \text{Then substitute the vertices into the Profit function: }\;P \:=\:0.50x + 0.80y\)
. . \(\displaystyle \text{and determine which vertex produces maximum profit.}\)

Thank you so much for all the help Soroban. When you say " must find the vertices of the region" do you mean something like this -

This is graphical method!

However, your graph is not correct. Follow Soroban's derivation carefully (including his definition of x & y). His graph is correct.
However, in your graph you are missing the line y ? 300