a rectangular piece of paper, measuring 50cm by 30cm, has a small square of side length X cut from each corner. The flaps are then folded up to form an open box in the shape of a cuboid.

a.) show the volume V of the cuboid may be expressed as V= 1500x - 160X^2 + 4x^3

b.) find the value X such that the cuboid has a maximum volume

c.) find the maximum possible volume of the cuboid

a.) v = x (50 - 2 x)(30 - 2 x) = x (1500 - 160 x + 4 x^2)

... = 1500 x - 160 x^2 + 4 x^3

b.) max for the quadratic ... -b / 2a = 160 / (2 * 4) = 20

c.) plug 20 into the volume equation

a.) Well, let's break it down, mathematically speaking, and find the volume of the cuboid. The length of the cuboid would be (50 - 2x) cm, the width would be (30 - 2x) cm, and the height would be x cm. Multiplying all these dimensions, we get:

V = (50 - 2x)(30 - 2x)(x)

Expanding that, we have:

V = (1500 - 100x - 60x + 4x^2)(x)
V = (1500 - 160x + 4x^2)(x)
V = 1500x - 160x^2 + 4x^3

And boom, there's our expression for the volume of the cuboid!

b.) To find the maximum volume, we need to find the value of X that gives us the highest value for the expression 1500x - 160x^2 + 4x^3. To do this, we can take the derivative of the expression with respect to x, set it equal to zero, and solve for x.

c.) To find the maximum possible volume of the cuboid, we substitute the value we found for X into the expression V = 1500x - 160x^2 + 4x^3. This will give us the maximum volume in terms of X.

a) To find the volume of the cuboid, we need to calculate the length, width, and height of the box.

Let's start by finding the length:
The original length of the paper is 50cm, and we cut X cm from each corner, so the length of the cuboid will be (50 - 2X).

Next, let's find the width:
The original width of the paper is 30cm, and we cut X cm from each corner, so the width of the cuboid will be (30 - 2X).

Finally, let's find the height:
The height of the cuboid is equal to the size of the square flaps we folded up, which is X cm.

Now we can calculate the volume of the cuboid:
Volume (V) = Length x Width x Height
V = (50 - 2X) x (30 - 2X) x X
V = (1500 - 100X - 60X + 4X^2) x X
V = (1500 - 160X + 4X^2) x X
V = 1500X - 160X^2 + 4X^3

Therefore, the volume of the cuboid can be expressed as V = 1500X - 160X^2 + 4X^3.

b) To find the value of X that gives the maximum volume of the cuboid, we can take the derivative of the volume equation and set it equal to zero.

dV/dX = 1500 - 320X + 12X^2

Setting dV/dX = 0:
0 = 1500 - 320X + 12X^2

Simplifying the equation:
12X^2 - 320X + 1500 = 0

Now, we can solve this quadratic equation either by factoring, using the quadratic formula, or completing the square. Let's use the quadratic formula:
X = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values:
X = (-(-320) ± √((-320)^2 - 4 * 12 * 1500)) / (2 * 12)

Simplifying:
X = (320 ± √(102400 - 72000)) / 24
X = (320 ± √(30400)) / 24
X = (320 ± 2√(7600)) / 24
X = (320 ± 2 * 2√(1900)) / 24
X = (320 ± 4√(1900)) / 24
X = (80 ± √(1900)) / 6

Therefore, the values of X that give the maximum volume of the cuboid are approximately:
X ≈ 26.17 cm (rounded to two decimal places) or X ≈ 2.16 cm (rounded to two decimal places).

c) To find the maximum possible volume of the cuboid, we can substitute the value of X into the volume equation.

For X ≈ 26.17 cm:
V ≈ 1500(26.17) - 160(26.17)^2 + 4(26.17)^3
V ≈ 39255.45 cm³ (rounded to two decimal places)

For X ≈ 2.16 cm:
V ≈ 1500(2.16) - 160(2.16)^2 + 4(2.16)^3
V ≈ 1407.06 cm³ (rounded to two decimal places)

Therefore, the maximum possible volume of the cuboid is approximately:
V ≈ 39255.45 cm³ (rounded to two decimal places) when X ≈ 26.17 cm.

To find the volume of the cuboid, we need to multiply its length, width, and height. Let's break down the process step by step:

a.) Show the volume V of the cuboid may be expressed as V = 1500x - 160X^2 + 4x^3

1) Start with a rectangular piece of paper measuring 50cm by 30cm.
2) Cut out squares with side length X from each corner.
3) After folding up the flaps, the resulting shape will have a length of (50 - 2X) cm, a width of (30 - 2X) cm, and a height of X cm.
4) Therefore, the volume of the cuboid is given by V = (50 - 2X) * (30 - 2X) * X.
5) Expanding this equation, we get V = (1500 - 100X + 4X^2) * X.
6) Simplifying further, V = 1500X - 100X^2 + 4X^3, which matches the expression mentioned.

b.) To find the value of X that gives the cuboid maximum volume, we need to find the critical points. These occur when the derivative of the volume equation is equal to zero.

1) Take the derivative of the volume equation with respect to X.
dV/dX = 1500 - 200X + 12X^2.

2) Set the derivative equal to zero and solve for X:
1500 - 200X + 12X^2 = 0.

3) This equation is a quadratic equation. You can solve it by factoring, completing the square, or using the quadratic formula.

c.) Once you find the critical values for X, you need to determine which one gives the maximum volume. This can be done by evaluating the second derivative of the volume equation.

1) Take the derivative of the volume equation again with respect to X.
d^2V/dX^2 = - 200 + 24X.

2) Evaluate the second derivative at each critical point found in step b.
Let's call the critical points X₁ and X₂.

If d^2V/dX^2 > 0 at X = X₁, then it's a minimum point.
If d^2V/dX^2 < 0 at X = X₁, then it's a maximum point.

Repeat this for X = X₂.

3) Compare the values. The X value that yields a maximum volume is the one for which the second derivative is negative.

4) Plug that X value into the cuboid volume equation V = 1500X - 100X^2 + 4X^3 to calculate the maximum possible volume.