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Answer :
- Define the volume of the box as a function of the cut size: $V(x) = x(20-2x)(12-2x)$.
- Find the critical points by solving $V'(x) = 12x^2 - 128x + 240 = 0$, which simplifies to $3x^2 - 32x + 60 = 0$.
- Determine the valid critical point within the domain $0 < x < 6$, which is $x = \frac{16 - 2\sqrt{19}}{3} \approx 2.427$.
- Calculate the dimensions of the box: length $\approx 15.145$ inches, width $\approx 7.145$ inches, and height $\approx 2.427$ inches. The height of the box is $\boxed{2.427}$ inches and the dimensions of the bottom of the box are $\boxed{15.145}$ inches by $\boxed{7.145}$ inches.
### Explanation
1. Problem Setup
We are given a 12-inch by 20-inch piece of cardboard, and we want to create an open box by cutting out squares of equal size from each corner and folding up the sides. Our goal is to find the dimensions of the box that maximize its volume. Let's denote the side length of the squares cut from the corners as $x$.
2. Expressing Volume as a Function
When we cut out the squares and fold up the sides, the dimensions of the base of the box will be $(20 - 2x)$ inches and $(12 - 2x)$ inches, and the height of the box will be $x$ inches. The volume $V$ of the box can be expressed as a function of $x$:
$$V(x) = x(20 - 2x)(12 - 2x)$$
Since the dimensions must be positive, we have the constraints: $x > 0$, $20 - 2x > 0$, and $12 - 2x > 0$. This implies $0 < x < 6$.
3. Finding Critical Points
Now, let's expand the expression for $V(x)$:
$$V(x) = x(240 - 40x - 24x + 4x^2) = 4x^3 - 64x^2 + 240x$$
To find the maximum volume, we need to find the critical points by taking the derivative of $V(x)$ with respect to $x$ and setting it equal to zero:
$$V'(x) = 12x^2 - 128x + 240 = 0$$
We can simplify this quadratic equation by dividing by 4:
$$3x^2 - 32x + 60 = 0$$
4. Solving for x
To solve the quadratic equation $3x^2 - 32x + 60 = 0$, we can use the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{32 \pm \sqrt{(-32)^2 - 4(3)(60)}}{2(3)} = \frac{32 \pm \sqrt{1024 - 720}}{6} = \frac{32 \pm \sqrt{304}}{6} = \frac{32 \pm 4\sqrt{19}}{6} = \frac{16 \pm 2\sqrt{19}}{3}$$
So the two possible values for $x$ are:
$$x_1 = \frac{16 + 2\sqrt{19}}{3} \approx 8.24$$
$$x_2 = \frac{16 - 2\sqrt{19}}{3} \approx 2.43$$
Since $0 < x < 6$, $x_1$ is outside the domain, so we consider $x_2 = \frac{16 - 2\sqrt{19}}{3}$.
5. Verifying Maximum
To confirm that $x_2$ corresponds to a maximum, we can take the second derivative of $V(x)$:
$$V''(x) = 24x - 128$$
Evaluate $V''(x_2)$:
$$V''\left(\frac{16 - 2\sqrt{19}}{3}\right) = 24\left(\frac{16 - 2\sqrt{19}}{3}\right) - 128 = 8(16 - 2\sqrt{19}) - 128 = 128 - 16\sqrt{19} - 128 = -16\sqrt{19} < 0$$
Since $V''(x_2) < 0$, $x_2$ corresponds to a local maximum.
6. Calculating Dimensions
Now we can calculate the dimensions of the box:
Height: $x = \frac{16 - 2\sqrt{19}}{3} \approx 2.427$ inches
Length: $20 - 2x = 20 - 2\left(\frac{16 - 2\sqrt{19}}{3}\right) = \frac{60 - 32 + 4\sqrt{19}}{3} = \frac{28 + 4\sqrt{19}}{3} \approx 15.145$ inches
Width: $12 - 2x = 12 - 2\left(\frac{16 - 2\sqrt{19}}{3}\right) = \frac{36 - 32 + 4\sqrt{19}}{3} = \frac{4 + 4\sqrt{19}}{3} \approx 7.145$ inches
7. Final Answer
Therefore, the dimensions of the box that maximize the volume are approximately:
Dimensions of the bottom of the box: $15.145$ inches by $7.145$ inches
Height of the box: $2.427$ inches
### Examples
Understanding optimization problems like this one is crucial in various real-world applications. For instance, in manufacturing, determining the optimal dimensions of a container to maximize volume while minimizing material usage can lead to significant cost savings. Similarly, in logistics, optimizing package sizes can improve storage and transportation efficiency. These principles extend beyond physical objects; they can also apply to resource allocation, such as maximizing profit within budget constraints or minimizing risk in investment portfolios.
- Find the critical points by solving $V'(x) = 12x^2 - 128x + 240 = 0$, which simplifies to $3x^2 - 32x + 60 = 0$.
- Determine the valid critical point within the domain $0 < x < 6$, which is $x = \frac{16 - 2\sqrt{19}}{3} \approx 2.427$.
- Calculate the dimensions of the box: length $\approx 15.145$ inches, width $\approx 7.145$ inches, and height $\approx 2.427$ inches. The height of the box is $\boxed{2.427}$ inches and the dimensions of the bottom of the box are $\boxed{15.145}$ inches by $\boxed{7.145}$ inches.
### Explanation
1. Problem Setup
We are given a 12-inch by 20-inch piece of cardboard, and we want to create an open box by cutting out squares of equal size from each corner and folding up the sides. Our goal is to find the dimensions of the box that maximize its volume. Let's denote the side length of the squares cut from the corners as $x$.
2. Expressing Volume as a Function
When we cut out the squares and fold up the sides, the dimensions of the base of the box will be $(20 - 2x)$ inches and $(12 - 2x)$ inches, and the height of the box will be $x$ inches. The volume $V$ of the box can be expressed as a function of $x$:
$$V(x) = x(20 - 2x)(12 - 2x)$$
Since the dimensions must be positive, we have the constraints: $x > 0$, $20 - 2x > 0$, and $12 - 2x > 0$. This implies $0 < x < 6$.
3. Finding Critical Points
Now, let's expand the expression for $V(x)$:
$$V(x) = x(240 - 40x - 24x + 4x^2) = 4x^3 - 64x^2 + 240x$$
To find the maximum volume, we need to find the critical points by taking the derivative of $V(x)$ with respect to $x$ and setting it equal to zero:
$$V'(x) = 12x^2 - 128x + 240 = 0$$
We can simplify this quadratic equation by dividing by 4:
$$3x^2 - 32x + 60 = 0$$
4. Solving for x
To solve the quadratic equation $3x^2 - 32x + 60 = 0$, we can use the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{32 \pm \sqrt{(-32)^2 - 4(3)(60)}}{2(3)} = \frac{32 \pm \sqrt{1024 - 720}}{6} = \frac{32 \pm \sqrt{304}}{6} = \frac{32 \pm 4\sqrt{19}}{6} = \frac{16 \pm 2\sqrt{19}}{3}$$
So the two possible values for $x$ are:
$$x_1 = \frac{16 + 2\sqrt{19}}{3} \approx 8.24$$
$$x_2 = \frac{16 - 2\sqrt{19}}{3} \approx 2.43$$
Since $0 < x < 6$, $x_1$ is outside the domain, so we consider $x_2 = \frac{16 - 2\sqrt{19}}{3}$.
5. Verifying Maximum
To confirm that $x_2$ corresponds to a maximum, we can take the second derivative of $V(x)$:
$$V''(x) = 24x - 128$$
Evaluate $V''(x_2)$:
$$V''\left(\frac{16 - 2\sqrt{19}}{3}\right) = 24\left(\frac{16 - 2\sqrt{19}}{3}\right) - 128 = 8(16 - 2\sqrt{19}) - 128 = 128 - 16\sqrt{19} - 128 = -16\sqrt{19} < 0$$
Since $V''(x_2) < 0$, $x_2$ corresponds to a local maximum.
6. Calculating Dimensions
Now we can calculate the dimensions of the box:
Height: $x = \frac{16 - 2\sqrt{19}}{3} \approx 2.427$ inches
Length: $20 - 2x = 20 - 2\left(\frac{16 - 2\sqrt{19}}{3}\right) = \frac{60 - 32 + 4\sqrt{19}}{3} = \frac{28 + 4\sqrt{19}}{3} \approx 15.145$ inches
Width: $12 - 2x = 12 - 2\left(\frac{16 - 2\sqrt{19}}{3}\right) = \frac{36 - 32 + 4\sqrt{19}}{3} = \frac{4 + 4\sqrt{19}}{3} \approx 7.145$ inches
7. Final Answer
Therefore, the dimensions of the box that maximize the volume are approximately:
Dimensions of the bottom of the box: $15.145$ inches by $7.145$ inches
Height of the box: $2.427$ inches
### Examples
Understanding optimization problems like this one is crucial in various real-world applications. For instance, in manufacturing, determining the optimal dimensions of a container to maximize volume while minimizing material usage can lead to significant cost savings. Similarly, in logistics, optimizing package sizes can improve storage and transportation efficiency. These principles extend beyond physical objects; they can also apply to resource allocation, such as maximizing profit within budget constraints or minimizing risk in investment portfolios.
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