Question

Quiz 6 section 4 The sum of the dimensions of the closed rectangular box with maximum volume that can be inscribed in the sphere 17 -24 012 062 022 002

Answer 1

Restriction: 914,7,Z) = x++y+z2-24 = 0 Volume of closed rectangular hox is - V=8eyz [o volume of ubold is lbh] now, TV=8y zł toxy h taky k Tg = 2x1 +2y ↑ +22F method of lagrange multiplier, we have", - T = nog 8 (y zł tx zq+xy R ) = x (2 x C + 2yh + 2zF) 892 = 2ax - a XZ= 204 - exy = 202 - from / 2 Y = ys=ge? = tx and from a 6 zatre from © am) 4² = 22 2 y = 12 2020-4-15 23:55

So, wyzz? Returning to constraint equation - g(x, y, z)= 0, we obtain 3 x² = 24 = x2=8 x= 282 which means x=y=2= 252 So, dimension x=y=2=252 manimize volume. and sum of dimension is x+y+z = 252 +252 +212 = (2+2+2) 2 - 652 Hence e option © is correct. 2020-4-15 23:56