求解MATHS問題,關於DIFFERENTIATION~!!

6.
The side of the square base = x mm = u m
where x = 1000/u

A (m²) : surface area of theblock

Height of the block
= 125/u²

A = 2*u² + 4*u*(125/u²)
A = 2u² + (500/u)
dA/du = 4u - (500/u²)
dA/du = 4(u³ - 125)/u²

d²A/du² = 4 + (1000/u³)

When u = 5, dA/du = 0, and d²A/du = 4 + (1000/5³) > 0
Hence, A is the minimum when u = 5
Minimum surface area = [2*5² + (500/5)] m³ = 150 m³


7.
y = (x - 1)(x - 2)
y = x² - 3x + 2

dy/dx = 2x - 3

Slope of the tangent at (0, 2)
= dy/dx | (x = 0)
= -3

Equation of the tangent at (0, 2)
y - 2 = (-3)(x - 0)
3x+ y - 2 = 0

Slope of the normal at (0, 2) = 1/3

Equation of the normal at (0, 2) :
y - 2 = (1/3)(x - 0)
x -3y + 6 = 0


8.
V : volume of the pyramid
H : height of the pyramid

Thesurface of the right pyramid is formed by the square base and 4 identicalisosceles triangles.
Let h m be the height of each isosceles triangle.

Total internal surface area :
x² +4 * [(1/2) * x * h] = 32
2xh = 32 - x²
h = (16/x) - (x/2)

By Pythagorean theorem :
H² +(x/2)² =h²
H² +(x/2)² =[(16/x) - (x/2)]²
H² +(x²/4)= (256/x²) -16 + (x²/4)
H² =(256/x²) -(16x²/x²)
H² =(16/x²)(16- x²)
H = (4/x)√(16 - x²)

Volume
= (1/3) * x² *H m³
= (1/3) * x² *(4/x)√(16 - x²) m³
= 4x√(16 - x²)/3 m³

V = 4x√(16 - x²)/3

dV/dx
= (4x/3)*[-x/√(16 - x²)] + (4/3)√(16 - x²)
= [(4/3)(16 - x²) -(4x²/3)]/ √(16 - x²)
= (4/3)[16 - x² - x²] /√(16 - x²)
= 8(8 - x²) /3√(16 - x²)

d²V/dx²
= [3(-16x)√(16 - x²) - (64 - 8x²)/3√(16 - x²)] / 9(16 - x²)
= [9(-16x)(16 - x²) -(64 - 8x)] / 27(16 - x²)^(3/2)
= (144x³ -2296x² - 64)/ 27(16 - x²)^(3/2)

When x = √8 : dV/dx = 0 and d²V/dx² ≈ 24.8 > 0
Hence, V is the maximum when x = √8
The maximum V = 4 *(√8) * √(16 - 8)/3 = 32/3
The maximum volume = 32/3 m³