Math Problem Statement

You have 4320043200 cm2 of cardboard to make a box with a square base and an open top. What is the largest possible volume of such a box?

Answer:  cm3

Solution

To solve this problem, we need to maximize the volume of the box given the constraints. Let's go step by step:

Variables and Formulae

  1. Let the side length of the square base be xx (in cm), and the height of the box be hh (in cm).

  2. The surface area of the box, which is given, consists of:

    • The square base (x2x^2).
    • Four vertical sides (4xh4 \cdot x \cdot h).

    Hence, the total surface area constraint is: x2+4xh=4320043200x^2 + 4xh = 4320043200

  3. The volume of the box is given by: V=x2hV = x^2 \cdot h

Express hh in Terms of xx

From the surface area constraint: h=4320043200x24xh = \frac{4320043200 - x^2}{4x}

Substitute hh into the Volume Formula

V=x24320043200x24xV = x^2 \cdot \frac{4320043200 - x^2}{4x} Simplify: V=x(4320043200x2)4V = \frac{x \cdot (4320043200 - x^2)}{4} V=4320043200xx34V = \frac{4320043200x - x^3}{4}

Maximize the Volume

To find the maximum volume, differentiate VV with respect to xx, and set the derivative equal to zero: dVdx=14(43200432003x2)\frac{dV}{dx} = \frac{1}{4} \cdot (4320043200 - 3x^2) 0=43200432003x20 = 4320043200 - 3x^2 3x2=43200432003x^2 = 4320043200 x2=1440014400x^2 = 1440014400 x=144001440037947.59cmx = \sqrt{1440014400} \approx 37947.59 \, \text{cm}

Solve for hh

Substitute xx back into the formula for hh: h=4320043200x24xh = \frac{4320043200 - x^2}{4x} h=43200432001440014400437947.59h = \frac{4320043200 - 1440014400}{4 \cdot 37947.59} h2880028800151790.3618963.43cmh \approx \frac{2880028800}{151790.36} \approx 18963.43 \, \text{cm}

Calculate the Maximum Volume

Substitute xx and hh into the volume formula: V=x2hV = x^2 \cdot h V=144001440018963.432.730×1013cm3V = 1440014400 \cdot 18963.43 \approx 2.730 \times 10^{13} \, \text{cm}^3

Final Answer:

2.73×1013cm3\boxed{2.73 \times 10^{13} \, \text{cm}^3}

Would you like detailed derivations or clarifications on specific steps?

Here are five related questions to expand the concept:

  1. How does adding a top to the box change the optimization problem?
  2. What happens to the volume if the surface area is doubled?
  3. Can you derive the maximum volume of a box with a rectangular base instead of a square base?
  4. How would the solution change if the box must have a height equal to its base side length?
  5. How does calculus help in solving real-world optimization problems like this one?

Tip: Always check the critical points in optimization problems for feasibility under given constraints.

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Math Problem Analysis

Mathematical Concepts

Calculus (Optimization)
Surface Area
Volume of a Box
Differentiation

Formulas

Surface Area: x² + 4xh = 4320043200
Volume: V = x²h
Height: h = (4320043200 - x²) / (4x)
Derivative of Volume: dV/dx = (4320043200 - 3x²) / 4

Theorems

Critical Points in Optimization
First Derivative Test for Maxima and Minima

Suitable Grade Level

Grades 11-12, College