Ethem S. answered 10/12/19
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a. Each bass will cost him $900 and each guitar will cost him $750. He buys b basses and g guitars. Write an equation for the total number of dollars, d, he will have to spend.
d = $900*b + $750*g
b. Jake has certain restrictions on the numbers of each kind of instrument he can stock: Write a system of inequalities expressing these restrictions.
i. His store is small, so he can buy no more than 50 instruments, total.
b+g <= 50
ii. Because guitars are more popular than basses, the number of guitars must be at least twice the number of basses.
g >= 2*b
iii. To get started, he must buy at least 17 guitars and at least 5 basses.
b >= 5
g >= 17
c. Plot the graph of the system in part b.
Draw line b+g=50. Then mark the region b+g <= 50 (Hint: set b and g to 0. 0+0 <= 50, which is true. Then point (0,0) must be in the region b+g <= 50)
Draw line g=2*b. Then mark region g >= 2*b (Hint: set b and g to 0. 0 >= 2*0, 0 >= 0, whith is true. Then point (0,0) must be in the region g >= 2*b)
Draw line b=5. Then mark region b >= 5
Draw line g = 17. Then mark region g >= 17
The intersection of all these regions is the feasible region.
d. Write an inequality state that Jake spends at least $36,000 on instruments. Darken the part of the feasible region in part c for which Jake spends at least $36,000.
From part a, d = $900*b + $750*g
36000 <= 900*b + 750*g
The boundary is the line 36000 = 900*b + 750*g
3600 = 90*b + 75*g (divide both sides by 5)
720 = 18*b + 15*g (divide both sides by 3)
240 = 6*b + 5*g. Draw this line. Then set b and g to zero => 36000 <= 0, which is not true. Then point (0,0) is not in the region.
e. What number of basses and what number of guitars would give the maximum feasible cost. What would this cost be?
From the graph, you can see that the maximum number of b and g is the intersection of b+g=50 and g = 2*b. Let's solve for this:
b+g=50
g=2*b
Substitute g = 2*b in the first equation:
b + (2*b) = 50
3*b = 50
b = 50/3 = 16.66666666.
g = 33.3333333
The solution has to satisfy
b>=5
g>=17
b+g<=50
g >= 2*b
Then b=16 and g=33 is one solution. Then the total cost is 900*b+750*g = 900*16 + 750*33 = $39150