A box with a square base and open top must have a
volume of . We wish to find the dimensions
of the box that minimize the amount of material used.
First, find a formula for the surface area of the box in
terms of only , the length of one side of the square
base.
[Hint: use the volume formula to express the height of
the box in terms of .]
Simplify your formula as much as possible.

Next, find the derivative, .

Now, calculate when the derivative equals zero, that is
when . [Hint: multiply both sides by .]
when We next have to make sure that this value of gives a
minimum value for the surface area. Let’s use the
second derivative test. Find .

Evaluate at the -value you gave above.

Ask by Maxwell Gough. in the United States

Mar 26,2025

Question

A box with a square base and open top must have a
volume of . We wish to find the dimensions
of the box that minimize the amount of material used.
First, find a formula for the surface area of the box in
terms of only , the length of one side of the square
base.
[Hint: use the volume formula to express the height of
the box in terms of .]
Simplify your formula as much as possible.

Next, find the derivative, .

Now, calculate when the derivative equals zero, that is
when . [Hint: multiply both sides by .]
when We next have to make sure that this value of gives a
minimum value for the surface area. Let’s use the
second derivative test. Find .

Evaluate at the -value you gave above.

Ask by Maxwell Gough. in the United States

Mar 26,2025

UpStudy AI · Accepted Answer

**Step 1. Express the Height in Terms of $$ x $$**

The volume of the box is given by  
$$
x^2 \cdot h = 364500.
$$  
Solving for $$ h $$, we have  
$$
h = \frac{364500}{x^2}.
$$

**Step 2. Write the Surface Area as a Function of $$ x $$**

Since the box has an open top, the surface area consists of the area of the base and the areas of the four sides. The area of the base is $$ x^2 $$ and the area of the four sides is  
$$
4 \times (x \cdot h) = 4xh.
$$  
Substitute the expression for $$ h $$:  
$$
A(x) = x^2 + 4x\left(\frac{364500}{x^2}\right) = x^2 + \frac{4 \cdot 364500}{x}.
$$  
Simplify the constant:  
$$
4 \cdot 364500 = 1458000.
$$  
Thus, the surface area in terms of $$ x $$ is  
$$
A(x) = x^2 + \frac{1458000}{x}.
$$

**Step 3. Find the First Derivative $$ A'(x) $$**

Differentiate $$ A(x) $$ with respect to $$ x $$:  
$$
A(x) = x^2 + \frac{1458000}{x} = x^2 + 1458000 \cdot x^{-1}.
$$  
Then,  
$$
A'(x) = 2x - 1458000 \cdot x^{-2} = 2x - \frac{1458000}{x^2}.
$$

**Step 4. Solve $$ A'(x)=0 $$ to Find Critical Points**

Set the derivative equal to zero:  
$$
2x - \frac{1458000}{x^2} = 0.
$$  
Multiply both sides by $$ x^2 $$ to eliminate the fraction:  
$$
2x^3 - 1458000 = 0.
$$  
Solve for $$ x^3 $$:  
$$
x^3 = \frac{1458000}{2} = 729000.
$$  
Take the cube root of both sides:  
$$
x = \sqrt[3]{729000}.
$$  
Notice that  
$$
729000 = 729 \times 1000, \quad \sqrt[3]{729}=9, \quad \sqrt[3]{1000}=10,
$$  
so  
$$
x = 9 \times 10 = 90.
$$

**Step 5. Compute the Second Derivative $$ A''(x) $$ for the Second Derivative Test**

Differentiate $$ A'(x) = 2x - 1458000 \cdot x^{-2} $$:  
$$
A''(x) = 2 + 2 \cdot 1458000 \cdot x^{-3} = 2 + \frac{2916000}{x^3}.
$$

**Step 6. Evaluate $$ A''(x) $$ at $$ x = 90 $$**

Substitute $$ x = 90 $$ into the second derivative:  
$$
A''(90) = 2 + \frac{2916000}{90^3}.
$$  
Since  
$$
90^3 = 90 \times 90 \times 90 = 729000,
$$  
we have  
$$
A''(90) = 2 + \frac{2916000}{729000} = 2 + 4 = 6.
$$  
Since $$ A''(90) > 0 $$, the function has a minimum at $$ x = 90 $$.

**Final Answers:**

- The surface area in terms of $$ x $$ is  
  $$
  A(x)=x^2+\frac{1458000}{x}.
  $$
- The first derivative is  
  $$
  A'(x)=2x-\frac{1458000}{x^2}.
  $$
- Setting the derivative to zero gives  
  $$
  A'(x)=0 \text{ when } x=90.
  $$
- The second derivative is  
  $$
  A''(x)=2+\frac{2916000}{x^3},
  $$
  and evaluating at $$ x=90 $$ gives  
  $$
  A''(90)=6.
  $$
**Simplified Answers:** - **Surface Area Formula:** $$ A(x) = x^2 + \frac{1458000}{x} $$ - **First Derivative:** $$ A'(x) = 2x - \frac{1458000}{x^2} $$ - **Critical Point:** $$ A'(x) = 0 \text{ when } x = 90 $$ - **Second Derivative:** $$ A''(x) = 2 + \frac{2916000}{x^3} $$ - **Second Derivative at $$ x = 90 $$:** $$ A''(90) = 6 $$

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