A- maths(Differentiation)問題4

5.
(i)
Length of the box = 32 - 2x
Width of the box = 20 - 2x
Height of the box = x

Volume of the box
= Length * Width * Height
= (32 - 2x)(20 - 2x)x
= (640 - 40x - 64x + 4x²)x
= (4x² - 104x + 640)x
= 4x³ - 104x² + 640x
= 4(x³ - 26x² + 160x)


(ii)
V = 4(x³ - 26x² + 160x)
V' = 4(3x² - 52x + 160)
V" = 4(6x - 52)

Put V' = 0:
4(3x² - 52x + 160) = 0
(x - 4)(3x - 40) = 0
x = 4 or x = 40/3 (rejected for 2(40/3) > 20)

When x = 4:
V" = 4(6*4 - 52) < 0
Hence, maximum V at x = 4


(iii)
At x = 4, maximum V
= 4(4³ - 26*4² + 160*4)
= 1152