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junior college 1 | H2 Maths
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I need help with 2ii
Date Posted: 4 years ago
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Date Posted:
4 years ago
Actually no need to use calculator for second derivative
True, the numbers can be easily estimated
Or rather, simplifying it gives -4n√3. Since n > 0, 4n√3 > 0 and therefore -4n√3 < 0
V = 2n²x - 6nx² + 4x³
dV/dx = 2n² - 12nx + 12x²
For stationary value of V, dV/dx = 0
2n² - 12nx + 12x² = 0
6x² - 6nx + n² = 0
Using quadratic formula,
x = ( -b ± √(b² - 4ac) ) / 2a
x = ( -(-6n) ± √((-6n)² - 4(6)(n²)) ) / 2(6)
x = ( 6n ± √(36n² - 24n²) ) / 12
x = ( 6n ± √12n² ) / 12
x = ( 6n ± √(3 x 4n²) ) / 12
x = (6n ± 2n√3) /12
(we can square root 4n² to get 2n. No need to worry about ± sign here as n is positive, so 4n² is also positive and the square root of a positive is represented by the principal root. eg √4 = 2 and not -√2 )
x = ½n ± ¹⁄6 n√3
Now,
When x = ½n + ¹⁄6 n√3,
2x = n ± ⅓ n√3
Since n > 0, then ⅓ n√3 > 0
So n + ⅓ n√3 > n, n - ⅓ n√3 < n.
n + ⅓ n√3 > n implies that 2x > n. That is not possible as the the width of the box must be greater than the sum of the 2 sides of the squares. I.e n > 2x
So x = ½n + ¹⁄6 n√3 is rejected.
Therefore x = ½n - ¹⁄6 n√3
dV/dx = 2n² - 12nx + 12x²
So d²V/dx² = 24x - 12n
= 24(½n - ¹⁄6 n√3) - 12n
= 12n - 4n√3 - 12n
= -4n√3 < 0
(since n > 0, which means 4n√3 > 0)
So the minimum value of V is obtained when x = ½n - ¹⁄6 n√3
(Side note : we don't need a calculator at all to answer this question)
dV/dx = 2n² - 12nx + 12x²
For stationary value of V, dV/dx = 0
2n² - 12nx + 12x² = 0
6x² - 6nx + n² = 0
Using quadratic formula,
x = ( -b ± √(b² - 4ac) ) / 2a
x = ( -(-6n) ± √((-6n)² - 4(6)(n²)) ) / 2(6)
x = ( 6n ± √(36n² - 24n²) ) / 12
x = ( 6n ± √12n² ) / 12
x = ( 6n ± √(3 x 4n²) ) / 12
x = (6n ± 2n√3) /12
(we can square root 4n² to get 2n. No need to worry about ± sign here as n is positive, so 4n² is also positive and the square root of a positive is represented by the principal root. eg √4 = 2 and not -√2 )
x = ½n ± ¹⁄6 n√3
Now,
When x = ½n + ¹⁄6 n√3,
2x = n ± ⅓ n√3
Since n > 0, then ⅓ n√3 > 0
So n + ⅓ n√3 > n, n - ⅓ n√3 < n.
n + ⅓ n√3 > n implies that 2x > n. That is not possible as the the width of the box must be greater than the sum of the 2 sides of the squares. I.e n > 2x
So x = ½n + ¹⁄6 n√3 is rejected.
Therefore x = ½n - ¹⁄6 n√3
dV/dx = 2n² - 12nx + 12x²
So d²V/dx² = 24x - 12n
= 24(½n - ¹⁄6 n√3) - 12n
= 12n - 4n√3 - 12n
= -4n√3 < 0
(since n > 0, which means 4n√3 > 0)
So the minimum value of V is obtained when x = ½n - ¹⁄6 n√3
(Side note : we don't need a calculator at all to answer this question)
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