A box of rectangular base and an open top has a surface area 600 cm2.
If the height of the box is equal to its width, find the dimensions that
give a maximum volume.
Let x be the width and height of the box, and y be the length of the box. Then the surface area of the box is given by xy + 2x^2 = 600.
We want to maximize the volume of the box V = x^2y. We will do this by expressing y in terms of x using the surface area formula, and then finding the maximum of the resulting expression.
From the surface area formula, we have y = (600 - 2x^2) / x. Substituting this expression for y in the volume formula, we get:
V(x) = x^2 * (600 - 2x^2) / x
V(x) = 600x - 2x^3
To find the maximum of V(x), we take its derivative and set it equal to zero:
V'(x) = 600 - 6x^2
0 = 600 - 6x^2
x^2 = 100
x = 10
So the width and height of the box are both 10 cm. Now we can find the length of the box using the surface area formula:
10y + 2(100) = 600
y = 40
The dimensions of the box that give a maximum volume are 10 cm by 10 cm by 40 cm.
To find the dimensions that give a maximum volume, we need to first express the volume of the box in terms of a single variable. Let's assume the width of the box is 'w' and the height is 'h'.
From the given information, we know that the surface area of the box is 600 cm^2. The surface area of the box consists of the area of the base and the four sides.
The area of the base is given by w * w = w^2.
The area of the four sides can be calculated as follows: there are two sides of dimensions w × h, and two sides of dimensions h × w, so the total area of the four sides is 2wh + 2wh = 4wh.
Adding the area of the base and the area of the four sides, we get the equation: w^2 + 4wh = 600.
Now, let's express the volume of the box in terms of 'w' and 'h'. The volume of a rectangular box is given by V = lwh, where l is the length (which is equivalent to the width in this case).
Since the box is open at the top, its height is equal to its width, so h = w.
Substituting h = w into the equation, we have V = w * w * w = w^3.
We now have two equations:
w^2 + 4wh = 600 (equation 1) and V = w^3 (equation 2).
To find the dimensions that give a maximum volume, we need to maximize equation 2, while satisfying equation 1.
To proceed, we can solve equation 1 for h in terms of w:
w^2 + 4wh = 600
4wh = 600 - w^2
h = (600 - w^2) / (4w)
We can substitute this expression for h into equation 2 to get the volume in terms of 'w' only:
V = w^3
V = w * ((600 - w^2) / (4w))
V = (600w - w^3) / 4
To find the maximum volume, we can take the derivative of V with respect to w and set it equal to zero. Let's differentiate V with respect to w:
dV/dw = (600 - 3w^2) / 4
Setting this derivative equal to zero and solving for w:
(600 - 3w^2) / 4 = 0
600 - 3w^2 = 0
3w^2 = 600
w^2 = 200
w = sqrt(200) ≈ 14.14
So, the width of the box that gives a maximum volume is approximately 14.14 cm.
Since h = w, the height of the box is also 14.14 cm.
Therefore, the dimensions of the box that give a maximum volume are approximately 14.14 cm for both the width and height.