An open cardboard box is to be made by cutting squares of side x cm from each corner of a card of side 60cm and folding the resulting "flaps" up to form the box. Find the value of x that gives the box a maximum capacity.

dimension of the box is

60-2x by 60-2x by x , where x < 30

Vol = x(60-2x)^2
= 3600x - 240x^2 + 4x^3
d(Vol)/dx = 3600 - 480x + 12x^2
= 0 for a max/min of Vol

divide each term by 12
x^2 - 40x + 300 = 0
(x-10)(x-40) = 0
x = 10 or x = 40, but x<30

x = 10

cut out flaps of 10 cm by 10 cm

check:
if x = 10 , Vol = 10(40)^2 = 16000
if x = 10.1 , Vol = 10.1(39.8)^2 = 15998.8 which is < 16000
8
if x = 9.9 , Vol = 9.9(40.2)^2 = 15998.8 which is < 16000
My answer is very plausible

To find the value of x that gives the box a maximum capacity, we need to determine an expression for the volume of the box in terms of x and then find the maximum value of this expression.

Let's start by considering the dimensions of the box. By cutting squares of side x cm from each corner of a card of side 60 cm, the resulting length of the base of the box will be (60 - 2x) cm, and the resulting width will be (60 - 2x) cm since the length and width of the base will be equal. The height of the box will be x cm.

Now, we can calculate the volume of the box by multiplying the length, width, and height:

Volume = (60 - 2x) * (60 - 2x) * x
= x(60 - 2x)^2
= 4x(30 - x)^2

Since we want to find the value of x that gives the box a maximum capacity, we need to determine when the volume reaches its maximum. To do this, we can find the value of x that makes the derivative of the volume with respect to x equal to zero:

dV/dx = 4(30 - x)^2 - 4x(30 - x)
= 4(30 - x)((30 - x) - x)
= 4(30 - x)(30 - 2x)

Setting this derivative to zero and solving for x gives:

4(30 - x)(30 - 2x) = 0

Since this equation is a product of three factors, it will be equal to zero if any of the factors are equal to zero. So we have two possibilities:

1. 4 = 0 (which is not possible)
2. (30 - x)(30 - 2x) = 0

Solving the second equation:

Case 1: (30 - x) = 0
x = 30

Case 2: (30 - 2x) = 0
2x = 30
x = 15

So the two possible values of x are 30 and 15.

To determine which value gives the maximum volume, we can compare the volumes calculated from these values:

Volume (x = 30) = 4(30)(0)^2 = 0
Volume (x = 15) = 4(15)(30 - 15)^2 = 4(15)(15)^2 = 4(15)(225) = 13500

Therefore, the value of x that gives the box a maximum capacity is x = 15 cm.

To find the value of x that gives the box a maximum capacity, we need to optimize the volume of the box. The volume of a rectangular box can be calculated by multiplying its length, width, and height.

Let's first determine the dimensions of the box when the squares of side x cm are cut from each corner of the card.

When the squares are cut from each corner, the length of the resulting box will be (60 - 2x) cm (since two squares are being cut at both ends, reducing the length by 2x cm).
The width of the resulting box will also be (60 - 2x) cm.
The height of the resulting box will be x cm (since the height of each flap is x cm).

Therefore, the volume V(x) of the resulting box can be calculated as follows:
V(x) = (60 - 2x) * (60 - 2x) * x

Now, we need to find the value of x that maximizes the volume V(x).

Taking the derivative of V(x) with respect to x, we get:
V'(x) = 4x(900 - 120x + 4x^2)

Setting V'(x) equal to 0 and solving for x, we find the critical points:
4x(900 - 120x + 4x^2) = 0

This equation has three solutions: x = 0, x = 15, and x = 37.5.

To determine which of these values gives the maximum volume, we can use the second derivative test or evaluate V(x) at these critical points.

By evaluating V(x) at x = 0, x = 15, and x = 37.5, we find the volume of the box:
V(0) = 0 cm^3
V(15) ≈ 16200 cm^3
V(37.5) ≈ 16875 cm^3

Thus, out of the three options, x = 37.5 cm gives the box a maximum capacity, with a volume of approximately 16875 cm^3.