The dimension of rectangular sheet of cardboard are 16 𝑐𝑚 by 10 𝑐𝑚 . Equal squares of length 𝑥 𝑐𝑚 are cut away from the four corners of the cardboard. The remaining edges are folded to form a rectangular open box of volume 𝑉 𝑐𝑚3 .
(i) The volume 𝑉 in terms of 𝑥
(ii) The maximum volume of the box
height = x
length = 16 - 2 x
width = 10 - 2x
V = x (160 - 52 x + 4 x^2)
V= 4 x*3 - 52 x^2 + 160 x
dV/dx = 12 x^2 - 104 x + 160
= 3 x^2 - 26 x + 40
= (x-2)(3x-20)
x = 2 for zero
V= 4 * 8 - 52 * 4 + 160 * 2
V = x(16-2x)(10-2x) , 0 < x < 5
= x(160 - 52x + 4x^2)
= 160x - 52x^2 + 4x^3
dV =/dx = 160 - 104x + 12x^2 = 0
3x^2 - 26x + 40 = 0
solve for x, (hint : it factors)
keep the x that fits my domain stated above