A closed box has a square base of side x metres and it's height h metres the total surface area of the box is 24metres squared.
a) Find the expression of h in terms of x.
b) Hence find the value of x that would make the volume of the maximum.
a) The total surface area of the box is given by:
2(x^2) + 4(xh) = 24
x^2 + 2xh = 12
x^2 + 2xh - 12 = 0
Now we can use the quadratic formula to find the values of h in terms of x:
h = (-2x ± √(4x^2 + 48))/2
h = (-2x ± √(4(x^2 + 12)))/2
h = (-2x ± 2√(x^2 + 12))/2
h = -x ± √(x^2 + 12)
Therefore, the expression of h in terms of x is:
h = -x ± √(x^2 + 12)
b) To find the maximum volume of the box, we need to maximize the volume formula:
V = x^2h = x^2(-x ± √(x^2 + 12))
Now, we need to find the critical points by taking the derivative of V with respect to x and setting it equal to zero:
dV/dx = 2x(-x ± √(x^2 + 12)) + x^2(-1 ± 1/(2√(x^2 + 12))) = 0
Solving for x will give us the critical points. Then, we can determine which critical point corresponds to the maximum volume.
This is a more involved calculation and can be done by hand or using a graphing calculator to plot the function and find the maximum point.