You are given 1200 cm^2 of cardboard to make a box with a square base and an open top. Find the largest possible volume of the box.
let the height be y and each side of the base be x cm
Surface area
= SA
= x^2 + 4xy = 1200
y = (1200 - x^2)/(4x)
= 300/x - (1/4)x
V= x^2 y = x^2(300/x - (1/4)x )
= 300x - (1/4)x^3
dV/dx = 300 - (3/4)x^2
= 0 for a max of V
(3/4)x^2 = 300
x^2 = 1200/3 = 400
x = √400 = 20
largest volume = 300(20) - (1/4)(8000)
= 4000 cm^3
To find the largest possible volume of the box, we need to maximize the volume of the box given the constraint that the total surface area of the cardboard stays within 1200 cm^2.
Let's assume the side length of the square base of the box is x cm. Since the box has an open top, it will have a height h cm.
The surface area of the box consists of the area of the base (which is a square) and the area of the four sides. The area of the base is x * x = x^2 cm^2.
Since the box has four sides, each side will have an area of x * h cm^2. Thus, the total area of the four sides will be 4 * x * h = 4xh cm^2.
Adding the areas of the base and the four sides gives us the total surface area of the box: x^2 + 4xh = 1200 cm^2.
Now, let's solve this equation for h in terms of x: h = (1200 - x^2) / (4x).
To find the largest possible volume, we maximize the volume of the box, which is given by V = x^2h.
Substituting the value of h from the equation above, we get V = x^2((1200 - x^2) / (4x)).
Simplifying this equation, we have V = (x/4)(1200x - x^3).
To find the largest possible volume, we need to find the value of x that maximizes V. We can do this by finding the critical points of V.
Taking the derivative of V with respect to x, we get dV/dx = 300 - 3x^2.
Setting dV/dx equal to zero to find critical points, we have 300 - 3x^2 = 0.
Simplifying further, we get x^2 = 100.
Taking the square root of both sides, we find x = ±10.
Since we are looking for the length of a side, the negative value is not meaningful, so we take x = 10 cm.
Now, we can calculate the corresponding value of h using the equation h = (1200 - x^2) / (4x).
Substituting x = 10, we get h = (1200 - 100) / 40 = 28 cm.
Therefore, the largest possible volume of the box is V = x^2h = 10^2 * 28 = 2800 cm^3.