L * x * h = 288 so hx = 96
A = 2*L*x + 2*h * L + 2 *h*x = 2 (Lx + Lh + hx) = 6 x + 6 h + 192
A = 6 x + 576/x + 192
hmm
dA/dx = 6 - 576/x
= 0 for min or max
x = 576/6 = 96
h = 1
L = 3
well 96 * 1 * 3 = 288 so that checks
A closed cardboard box is designed to hold a volume of 288 cm^3 . The length is 3 cm and the width x cm .Show that the total surface area A cm^2 is given by A=6x^2+768/x and find the dimensions of the box which will make A a minimum. [8 Marks]
2 answers
w = x
l = 3
let the height be h
wlh = V
3xh = 288
xh = 96
h = 96/x
Area = 2(wl + lh + wh)
= 2(3x + 3(96/x) + x(96/x) )
= 6x + 576/x + 192
or (6x^2 + 576 + 192x)/x <==== which does not match what is given in the question
dA/dx = 6 - 576/x^2 = 0 for a minimum of A
576/x^2 = 6
6x^2 = 576
x^2 = 96
x = √96 = 4√6 or appr 9.8 cm
l = 3
let the height be h
wlh = V
3xh = 288
xh = 96
h = 96/x
Area = 2(wl + lh + wh)
= 2(3x + 3(96/x) + x(96/x) )
= 6x + 576/x + 192
or (6x^2 + 576 + 192x)/x <==== which does not match what is given in the question
dA/dx = 6 - 576/x^2 = 0 for a minimum of A
576/x^2 = 6
6x^2 = 576
x^2 = 96
x = √96 = 4√6 or appr 9.8 cm