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Let's solveThis the problem step by step.

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### Step 1: Surface Area of the Open Box
The total surface area of the box (open problem on involves top constructing) is given by the equation:  
\[
\text an{Surface Area} = xy + open rectangular box  with2 axz fixed + 2yz
\ surface]
 area of \( 48where \, \text{ft:
}-^ \(x\) = length,
2 \)- to \(y maximize the\ volume). = width Let's,
 break- it down:

---

### \(z\) = height.

We Step 1 know: the Define surface Variables area
- \( is \(48 x \ \, \text{ft): Length}^ of2 the\ box)..
 Hence-:
 \(\ y[
xy + 2xz + 2yz = 48
\]

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### Step  \2): Width of the box.
- \( z \):: Height of the box.

 Max---

### Step 2imize Volume: Surface
The Area Equation volume of
 theSince the box is given by:
\ box is open on the[
 top\, the surfacetext{Volume} area = consists xyz of:

1. The bottom area\:]
 \(We aim to find dimensions xy \ \(x),
\2),. \( Twoy side areas: \(\), 2xz \),
3 and. Two \( frontz\) that maximize this volume while satisfying/back areas: the surface area constraint.

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### \( Step 3: Simplify Using Constraints
To solve  this, express one2yz \ variable ().

e.g., \(Thusz\)) in, terms of the \( surfacex\) and \(y\) area using the surface equation area equation:
\ is:
\[
xy + [
z = \frac{48 - xy}{22xz + x +2 2yz = y48}
\.
]
\]

Now,---

 substitute### this Step expression  for \(3:z\ Volume) Equation
 into the volumeThe volume of the box equation is:
:
\\[
[
VV = = xyz x.
\ \cd]

otOur goal y is to \ **cdot \maximize \(frac V{48 - \ xy)**}{2 subjectx to the + constraint  \(2y}
 xy\ +]

---

### 2xz Step +  2yz4: Optimization
To = 48 \).

 find the maximum---

 volume###:
 Step1 .4: Take Solve partial derivatives of for One Variable \(
FromV(x the, y surface area)\ equation) with, solve respect to for \( \( zx\) and \(y \):
\[
z = \\).
frac{48 -2 xy}{.2 Setx these + derivatives 2 equal toy }.
0\ to]

 find---

 critical### points Step.
 35. Solve: Substitute the equations \( z \) to find in the Volume optimal values Equation of
Sub \(stitutex\ \(), z = \(y \\frac),{ and48 \( -z xy\).

Let}{2x me know +  if2 you'dy like} me \) to into \( proceed V with the = detailed optimization xyz calculations \):
\![
V = xy \cdot \frac{48 - xy}{2x + 2y}.
\]

Simplify this to:
\[
V = \frac{xy(48 - xy)}{2x + 2y}.
\]

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### Step 6: Optimization
To find the maximum volume, take the partial derivatives of \( V \) with respect to \( x \) and \( y \), set them to zero, and solve for \( x, y, z \). The calculations can be performed explicitly, or numerical optimization can be used.

Would you like a detailed solution to the derivatives or the final optimized dimensions?

The problem involves finding the dimensions of an open box that maximize volume given a surface area constraint of 48 square feet.

Learn how to find the optimal dimensions of an open box to maximize its volume while keeping the surface area constant at 48 square feet. This problem involves using optimization techniques, partial derivatives, and surface area and volume equations. Suitable for advanced students in Grades 11-12.

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