Example Problem: Find 2 non-negative numbers whose sum is 16 and so the product is a maximum. 1. a + b = 16 2. P = aXb 3. a + b = 16 b = 16 – a 4. P = a(16-a) 16a – a^2 Next you take the derivative and solve for a: 16 – 2a = 0 a = 8 Plug 8 into problem before you solved for the derivative 8 > 16(8) – (8)^2 = 64 Plug into primary. 64 = 8 X b a = 8, b = 8
Well, this has officially been the worst night of my life....so if this blog sucks o well...
Find two numbers whose sum is a max when the one times the first number and 3 times the second number are added and whose product is a 12. 1) Givens: a + 3b = max ab = 12 2) Primary: a + 3b = max 3) Secondary: ab = 12 4) Figure it out: a = 12/b -plug a into primary: 12/b + 3b = 0 -solve for b, then plug in the b-value and solve for a an you're done.
-solve for b then plug in your b value and solve for a and you should get
Find 2 non-negative numbers whose sum is 16 and so the product is a maximum. 1. a + b = 16 2. P = aXb 3. a + b = 16 b = 16 – a 4. P = a(16-a) 16a – a^2 Next you take the derivative and solve for a: 16 – 2a = 0 a = 8 Plug 8 into problem before you solved for the derivative 8 > 16(8) – (8)^2 = 64 Plug into primary. 64 = 8 X b a = 8, b = 8
Example Problem:
ReplyDeleteFind 2 non-negative numbers whose sum is 16 and so the product is a maximum.
1. a + b = 16
2. P = aXb
3. a + b = 16
b = 16 – a
4. P = a(16-a)
16a – a^2
Next you take the derivative and solve for a:
16 – 2a = 0
a = 8
Plug 8 into problem before you solved for the derivative
8 > 16(8) – (8)^2 = 64
Plug into primary.
64 = 8 X b
a = 8, b = 8
Find 2 positive numbers whose sum is 22 and whose product is a max.
ReplyDelete1. a + b = 22
2. primary = a x b
3. a + b = 22 a = 22 - b
4. p = (22 - b)b
22b - b^2
derive : 22 - 2b
set = 0
22 - 2b = 0
-2b = - 22
b = 11
5. Plug b ( 11 ) into problem
a + 11 = 22
a = 11
a = 11, b = 11
Two non negative numbers whose
ReplyDeletesum is 12 and the product is a maximum.
1. a+b=12 a x b=P
2. a x b= p--primary equation
3. a+b=12
in this step you have to solve for
either a or b in order to plug in
for one of the variables making you able
to solve the problem.
b=12-a
4. plug in 12-a for b.
a(12-a)=p
12a-a^2
derivative
12-2a=p
a=-6
12-2(6)=0=p
-6+b=12
b=18
a=-6 and b=18 is the final
answer.
Well, this has officially been the worst night of my life....so if this blog sucks o well...
ReplyDeleteFind two numbers whose sum is a max when the one times the first number and 3 times the second number are added and whose product is a 12.
1) Givens: a + 3b = max ab = 12
2) Primary: a + 3b = max
3) Secondary: ab = 12
4) Figure it out:
a = 12/b
-plug a into primary: 12/b + 3b = 0
-solve for b, then plug in the b-value and solve for a an you're done.
-solve for b then plug in your b value and solve for a and you should get
the sum of two numbers is ten and the product is a max.
ReplyDelete10=a+b
p=ab
a+b=10 a=b-10
p=b-10(b)
b^2-10b
2b-10=0
2b=10
b=5, a=5
Find 2 non-negative numbers whose sum is 16 and so the product is a maximum.
ReplyDelete1. a + b = 16
2. P = aXb
3. a + b = 16
b = 16 – a
4. P = a(16-a)
16a – a^2
Next you take the derivative and solve for a:
16 – 2a = 0
a = 8
Plug 8 into problem before you solved for the derivative
8 > 16(8) – (8)^2 = 64
Plug into primary.
64 = 8 X b
a = 8, b = 8