application of derivatives:
an open box is to be made by cutting small congruent squares from corners of a 12cm by 12cm . sheet of tin and bending up the sides . how large should the squares cut from the corners to be make the box hold as much as possible ?
To find the size of squares that should be cut to maximize the volume of the box, we need to apply the concept of derivatives.
Let's start by visualizing the problem. Suppose we cut x cm squares from each corner of the tin sheet. Now, when we fold up the sides, an open box is created.
To determine the dimensions of the box, let's consider the length, width, and height.
The length of the box will be 12 cm - 2x (since squares are cut from both ends), and similarly, the width will also be 12 cm - 2x. The height of the box will be x cm.
Now, the volume of the box can be calculated using the formula: V = length × width × height.
V = (12 - 2x) × (12 - 2x) × x
V = x(12 - 2x)(12 - 2x)
To maximize the volume, we need to find the value of x that maximizes the function V(x).
To do this, we can differentiate V(x) with respect to x:
dV/dx = 12(12 - 2x) + x(-2)(12 - 2x) + x(12 - 2x)(-2)
dV/dx = 144 - 24x - 24x + 4x^2 - 24x + 4x^2
Simplifying, we get:
dV/dx = 8x^2 - 72x + 144
Now, we can find the critical points by setting dV/dx = 0 and solving for x:
8x^2 - 72x + 144 = 0
Dividing by 8, we get:
x^2 - 9x + 18 = 0
Factoring the quadratic equation, we have:
(x - 6)(x - 3) = 0
So, x = 6 or x = 3.
Now, we need to determine which value of x maximizes the volume. To do this, we can use the second derivative test or evaluate the function at these critical points.
V(x) = x(12 - 2x)(12 - 2x)
For x = 6:
V(6) = 6(12 - 2(6))(12 - 2(6))
= 6(12 - 12)(12 - 12)
= 0
For x = 3:
V(3) = 3(12 - 2(3))(12 - 2(3))
= 3(12 - 6)(12 - 6)
= 3(6)(6)
= 108 cm^3
Hence, cutting 3 cm squares from each corner will maximize the volume of the box, resulting in a volume of 108 cm^3.