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Quick Geometrical Application of Calculus Question? (1 Viewer)
- Thread starter Zokunu
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MrBeefJerky
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3b)
y'' = -2x
Integrating,
y' = -x^2 + C
When f'(x)=0 (min turning point) x= 3
0=-9 + C
C= 9
Therefore, y' = -x^2 +9
Integrating,
y= -x^3/3 +9x + C
When x=3, y = 7
7 = -9 + 27 + C
C = -11
Therefore, y=-x^3/3 + 9x - 11
Is that the correct answer? Tell me if I did something wrong
y'' = -2x
Integrating,
y' = -x^2 + C
When f'(x)=0 (min turning point) x= 3
0=-9 + C
C= 9
Therefore, y' = -x^2 +9
Integrating,
y= -x^3/3 +9x + C
When x=3, y = 7
7 = -9 + 27 + C
C = -11
Therefore, y=-x^3/3 + 9x - 11
Is that the correct answer? Tell me if I did something wrong
Last edited:
MrBeefJerky
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3c) xy=300
A=(x-5)(y-3)
A=(x-5)(300/x - 3)
= 300-3x-1500/x + 15
= 315 - 3x -1500/x
dA/dx = -3 + 1500/x^2
when dA/dx = 0
x = sqr(500)
= 10sqr(5)
y= (300/x - 3)
= (30/sqr 5 - 3)
To prove maximum, find d^2A/dx^2
d^2A/dx^2 = -3000/x^3
Sub x= 10sqr(5) and its a maximum
Therefore dimensions are x=10sqr(5) and y= (30/sqr5 - 3)?
A=(x-5)(y-3)
A=(x-5)(300/x - 3)
= 300-3x-1500/x + 15
= 315 - 3x -1500/x
dA/dx = -3 + 1500/x^2
when dA/dx = 0
x = sqr(500)
= 10sqr(5)
y= (300/x - 3)
= (30/sqr 5 - 3)
To prove maximum, find d^2A/dx^2
d^2A/dx^2 = -3000/x^3
Sub x= 10sqr(5) and its a maximum
Therefore dimensions are x=10sqr(5) and y= (30/sqr5 - 3)?
MrBeefJerky
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Didn't it say the dimensions of the printing area not the dimensions of the rectangular sheet of paper?
panda15
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AsianFrustratio
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Hmmm, if f''(x)=-2x, how can (3,7) be a minimum turning point? f''(3)=-6 < 0, so doesn't that mean that (3,7) must be a max?
I know this has nothing to do with your question (sorry!), but it was something interesting that I felt the need to point out.
I know this has nothing to do with your question (sorry!), but it was something interesting that I felt the need to point out.