Question 1067708
your objective function is p = 12x + 10y


p is the profit.
x is the number of cordless phones.
y is the number of landline phones.


your constraint equations are:


x >= 300
x + 3/2 * y <=1000
y >= 0


x >= 300 because the number 0of cordless phone has to be greater than or equal to 300.
y >= 0 because the number of landline phones has to be greater than or equal to 0.
x + 3/2 * y <= 100 because the total manufacturing hours have to be less than 1000 and it takes 1 hour to manufacture a cordless phone and 3/2 hours to manufacture a landline phone.


using the desmos calculator (www.desmos.com/calculator), you would graph the opposite inequalities and the unshaded region is your feaqsible region.


the corner points of your feasible region are the max/min values for your objective function.


your constraint equations are, once again:


x >= 300
x + 3/2 * y <=1000
y >= 0


you would graph:


x <= 300
x + 3/2 * y >=1000
y <= 0


your graph will look like this:


<img src = "http://theo.x10hosting.com/2017/021001.jpg" alt="$$$" </>


the corner points of your graph are at:


(300,466.667)
(300,0)
(1000,0)


your objective function of p = 12x + 10y yields the following profite at each of these corner points respectively.


8266.67
3600
12000


to maximize your profit, you need to manufacture 1000 cordless phones and 0 landline phones.


your constraints need to be met.
total hours to manufacture is less than or equal to 1000 (met).
number of cordles phones greater than or equal to 300 (met).
number of landline phones greater than or equal to 0 (met).


solution looks good.