Each hat uses 0.2 kilograms of yarn and each scarf uses 0.1 kilograms of yarn. She wants to make 3 times as many scarves as hats
1 answer:
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Answer:
Maximum Profit = 13050
when
The Toy company sells Jordan Size hoops = 7
The Toy company sells Shaq Size hoops = 3
Step-by-step explanation:
Given - The Toy Company sells two types of kids' basketball hoops, Jordan size and Shaq size. A neighborhood needs no more than 10 Jordan sizes and Shaq sizes all together. The Company requires the neighborhood to order at least 2 Jordan sizes and at least 3 Shaq sizes. The Toy Company makes a profit of $1,350 on the Jordan size and $1,200 on the Shaq size.
To find - What is the maximum profit they can make given the constraints above?
Proof -
Let us assume that,
The Toy company sells Jordan Size hoops = x
The Toy company sells Shaq Size hoops = y
Now,
Given that, A neighborhood needs no more than 10 Jordan sizes and Shaq sizes all together.
⇒x + y ≤ 10
Now,
Given that, The Company requires the neighborhood to order at least 2 Jordan sizes and at least 3 Shaq sizes.
⇒ x ≥ 2
y ≥ 3
Now,
Given that,
The Toy Company makes a profit of $1,350 on the Jordan size and $1,200 on the Shaq size.
So,
The objective function becomes
Z = 1350x + 1200 y
So,
The Linear Programming Problem (LPP) becomes
Maximize Z = 1350x + 1200 y
Subject to
x + y ≤ 10
x ≥ 2
y ≥ 3
x, y ≥ 0
We will Solve the LPP by Graphical method.
The graph is as follows :
The points on the Boundary are -
A(2, 8)
B(2, 3)
C(7, 3)
So,
Points (x,y) Objective function value ( Z = 1350x + 1200y)
B(2,3) 1350(2) + 1200(3) = 6300
C(7,3) 1350(7) + 1200(3) = 13050
A(2,8) 1350(2) + 1200(8) = 12300
So,
Maximum value = 13050 at point C(7,3)
∴ we get
Maximum Profit = 13050
when
The Toy company sells Jordan Size hoops = x = 7
The Toy company sells Shaq Size hoops = y = 3
Remark
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.