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A small company produces flower groups for Christmas. They have access to two sorts of flowers “hyacinter” and “Christmas stars”. For flower group A it’s needed 5 hyacinter and 2 Christmas stars and for group B it’s needed 3 hyacinter and 4 Christmas stars.

During a day the company have capacity to make 160 flower flower groups. They have daily access to 680 hyacinter and 500 Christmas stars. How many flower groups of each kind should be produced if one wants to have the maximum daily win? Every group A gives 20 dollars win and every group B 30 dollars win? Assume that the whole daily production is being sold.

Can somone help ?

What I tried : Let "x" be the number of groups A and "y" of B

Then 2x+4y=500 and 5x+3y=680

I need the maximum value of 20x+30y

Anything wrong ? If not, how should I continue ?

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    In what subject matter did this question arise, Pedro? We can better help you if you indicate the subject of the course you are studying, and your experience in math: the level at which you are studying. You see, this can be answered in many may ways, depending on the "tools" you've learned. We like to tailor answers to the appropriate level of your knowledge. – amWhy May 17 '20 at 14:33
  • You don't have equations but inequalities. $2x+4y\leq500$ for example. Also, you have omitted the constraint that the company can't make more than $160$ groups. P.S. "jacintos" is "hyacinths" in English. – saulspatz May 17 '20 at 14:34
  • @amWhy this exercise is an 11'th grade level – Pedro Alvarès May 17 '20 at 14:44

1 Answers1

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You are almost done.

During a day the company have capacity to make 160 flower groups

This is the constraint $x+y\leq 160$

You have the the LP

$x$: Number of group A

$y$: Number of group B

\begin{eqnarray} \\ \textrm{max} \ \ 20x+30y\\ 2x+4y\leq 500 \\ 5x+3y\leq 680 \\ x+y\leq 160 \\ x,y \geq 0 \end{eqnarray}

This LP can be solved with the simplex method or graphically. Here and here are examples, where the graphical method has been applied.

callculus42
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