Math Problem Statement

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

Solution

Let's solveThis the problem step by step.

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:

\ 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$$ --- ### Step \2): Width of the box. - $$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. --- ### \( 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}.$$ --- ### Step 6: Optimization To find the maximum volume, take the partial derivatives of $$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?

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Math Problem Analysis

Mathematical Concepts

Optimization
Surface Area
Volume
Calculus
Algebra

Formulas

Surface Area = xy + 2xz + 2yz
Volume = xyz
z = (48 - xy) / (2x + 2y)

Theorems

Optimization using partial derivatives
Constraints in optimization problems

Suitable Grade Level

Grades 11-12

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