a manufacturer desgins open top rectangular prism gift boxes that modelled by the function V=x^2h where top is square. The surface area of one box is 108cm^2 find the dimentsion of this box that would maxmise its volume
Let's denote the dimensions of the box as x, x, and h for the length, width, and height respectively. We are given that the surface area of the box is 108 cm^2, so we have:
2(x^2) + 4(xh) = 108
We are also given that the volume of the box is given by V = x^2h.
To find the dimensions that would maximize the volume, we need to find the values of x and h that satisfy the surface area equation and maximize the volume equation.
From the surface area equation, we can simplify it to:
x^2 + 2xh = 54
x(x + 2h) = 54
x = 54/(x + 2h)
Now, substitute x back into the volume equation:
V = (54/(x + 2h))^2 * h
V = 54^2 * h / (x + 2h)^2
To maximize the volume, we need to find dV/dx and set it to zero:
dV/dx = 0 = -54^2 * h * 2(x + 2h) / (x + 2h)^3
0 = -108h / (x + 2h)
Since h cannot be zero, the numerator must be zero:
108h = 0
h = 0
Since h cannot be zero, there is a mistake in the calculations. Let's correct it:
Let's substitute x = 54/(x + 2h) back into the surface area equation:
x^2 + 2xh = 54
(54/(x + 2h))^2 + 2(54/(x + 2h))h = 54
54^2 / (x + 2h)^2 + 2(54h) / (x + 2h) = 54
54^2 + 108h = 54(x + 2h)
54^2 + 108h = 54x + 108h
54^2 = 54x
54 = x
Now we have the value of x, we can find the value of h using the surface area equation:
x^2 + 2xh = 54
54^2 + 2(54)h = 54
54^2 + 108h = 54
108h = 54^2
h = 54
Therefore, the dimensions of the box that would maximize its volume are x = 54 cm, x = 54 cm, and h = 54 cm.