a rectangular box of given volume is to be "a" times long as it is wide. find the dimension for which it has t

Answer 1

haven't you forgotten to add something?

Answer 2

permit width = x length = ax height = V/(ax^2) permit floor section = S S = 2(lw + wh + lh) S = 2(ax^2 + V/x + V/ax) S = 2(ax^2 + V * x^(-a million) + V/a *x^(-a million)) dS/dx = 2(2ax - V * x^(-2) - V/a * x^(-2)) dS/dx = 4ax - 2V / x^2 -V/(ax^2) dS/dx = (4(a^2)(x^3) - 2aV - V) / (ax^2) dS/dx = 0 4(a^2)(x^3) - 2aV - V = 0 4(a^2)(x^3) = 2aV + V x^3 = (2aV + V) / (4a^2) x = 0.33 root of all that. because of fact the quantity is given and a is a persevering with, this might desire to be a suitable answer. to analyze that this value of x supply you the least achievable floor section, take the 2d by-fabricated from the S function, it relatively is two(2a + 2V * x^(-3) + 2V/a * x^(-3)). This function is beneficial for all beneficial values of x, and since the width of the container must be beneficial, it may supply a relative min for this reason (it relatively is, it supplies the least achievable value of S).

Answer 3

for which it has 'what'?