Verified answer

A) It's open so it doesn't have a top:

The inner SA = x * 2x + 2xh + 2 * 2xh

SA = 2x² + 6xh

volume V = x * 2x * h = 2x²h .... [eqn 1]

The internal SA = 12 units²

so 2x² + 6xh = 12

6xh = 12 - 2x²

h = (12 - 2x²) / 6x ... [eqn 2]

subs [eqn 2] into [eqn 1] → V = 2x² * [(12 - 2x²) / 6x]

V = (x/3) * 2(6 - x²)

V = (2x / 3) (6 - x²)

B) V = (2/3) * x(6 - x²)

V = (2/3) * [6x - x³]

dV/dx = 4 - 2x² and d²V/dx² = -4x

dV/dx = 0 when 4 - 2x² = 0

x² = 2

x = √2 ... rejecting the negative answer b/c length is positive

When x = 2, d²V/dx² < 0 ... so local max when x = √2

so the volume is a maximum when x = √2

a million) kx^2 +12x +ok=0 - if ok=0: we've: 12x=0 => x=0. - if ok different from is 0. => delta = 12^2 - 4.ok^2 = one hundred forty four - 4k^2 + if delta < 0 => one hundred forty four - 4k^2 <0 => ok> 6 or ok< - 6 => the equation do not have roots + if delta = 0 => ok = 6 or ok = -6 => the equivalent root: x = - 12/2k. + if delta >0 => ok>6=> the equivalent roots are x1 and x2. 2) x + y = 2 (a million) x^2 +2y =12 (2) from (a million)=> y =2 – x from (2)=> x^2 +2y=12 => x^2 +2(2-x) -12=0 ? x^2 +4 – 2x -12 = 0 => x^2 – 2x – 8=0 ? x= 4 and x= - 2 from (a million) => y= - 2 and y = 4 (x;y) = (4; - 2) and ( -2;4).