Story problem linear equation with multiple variables

[tklopfstein]

Hello,

I need help setting this problem up:

A sausage company makes two different kinds of hot dogs, regular and all beef. Each pound of all-beef hot dogs requires .75 lb of beef and 0.2 spices, each pound of regular hot dog requires .18 pound of beef, 0.3 pound of pork and 0.2 pound of spices. Suppliers can deliver at most 1020 pounds of beef, 600 pounds of pork and at least 500 pounds of spices. If the profit is $0.60 on each pound of beef hot dogs and $.40 on each pound of regular hot dogs, how many pounds of each should be produced to maximize profit?

I tried this with no luck:

Let x = the amount of all beef

Let y = the amount of regular
X+Y=2120
.75*1020+.20*500=1x
.18*1020+.30*600+.20*500=1y

I don't think this is the right way to set this up, but I'm just not sure how to set this thing up. Can anyone please help me.

Tammy K

 

[Denis]

Well, that's sure not clear, Tammy.

"Each pound of all-beef hot dogs requires .75 lb of beef and 0.2 spices"
.75 + .2 = .95 ; the bun weighs .05 lb :?:

"each pound of regular hot dog requires .18 pound of beef, 0.3 pound of pork and 0.2 pound of spices."
.18 + .3 + .2 = .68 ; heavy bun that weighs .32 lb :?:

Unless somebody else sees something I'm missing, all that can be done is:

Hotdog A : .75 beef + .00 pork + .20 spice : profit = .60 per hotdog
Hotdog B : .18 beef + .30 pork + .20 spice : profit = .40 per hotdog

At least 1020 lb of beef and at least 500 lb of pork MUST be purchased.

Since spice order = 600 lb at MOST, then 600/.20 = 3000 hotdogs as limit.
SO: if number of hotdogs A = a, then number of hotdogs B = 3000 - a.

I get this as maximum profit case:

--number--beef--pork---spice---profit----example of a calculation
A: 1333 999.75 000.00 266.6 $799.80 [1333 * .75 = 999.75]
B: 1667 300.06 500.10 333.4 $666.80 [1667 * .40 = 666.80]
===========================
3000 1299.81 500.10 600.0 $1466.60

Ok, before talking setups/formulas et al, let us know if that's the
way it all works...