Math Problem Statement

If 1500 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box

Solution

To find the largest possible volume of a box with a square base and an open top using 1500 square centimeters of material, we need to set up and solve an optimization problem.


Step 1: Define variables

  • Let xx be the length of one side of the square base.
  • Let hh be the height of the box.

Step 2: Constraints and relationships

  1. Surface area constraint: The box has:

    • A square base with an area x2x^2,
    • Four rectangular sides, each with an area xhxh.

    Total surface area is: x2+4xh=1500x^2 + 4xh = 1500

  2. Volume of the box: The volume VV is: V=x2hV = x^2h


Step 3: Solve for hh in terms of xx

From the surface area equation: h=1500x24xh = \frac{1500 - x^2}{4x}


Step 4: Substitute hh into the volume equation

V=x21500x24xV = x^2 \cdot \frac{1500 - x^2}{4x} Simplify: V=x(1500x2)4V = \frac{x(1500 - x^2)}{4} V=1500xx34V = \frac{1500x - x^3}{4}


Step 5: Maximize VV

Differentiate VV with respect to xx: dVdx=15003x24\frac{dV}{dx} = \frac{1500 - 3x^2}{4}

Set dVdx=0\frac{dV}{dx} = 0: 15003x2=01500 - 3x^2 = 0 x2=500x^2 = 500 x=500=10522.36cmx = \sqrt{500} = 10\sqrt{5} \approx 22.36 \, \text{cm}


Step 6: Determine hh

Substitute x=105x = 10\sqrt{5} into the equation for hh: h=1500(105)24105h = \frac{1500 - (10\sqrt{5})^2}{4 \cdot 10\sqrt{5}} h=1500500405h = \frac{1500 - 500}{40\sqrt{5}} h=1000405=255=55cmh = \frac{1000}{40\sqrt{5}} = \frac{25}{\sqrt{5}} = 5\sqrt{5} \, \text{cm}


Step 7: Find the maximum volume

V=x2h=(105)255V = x^2h = (10\sqrt{5})^2 \cdot 5\sqrt{5} V=50055=25005cm3V = 500 \cdot 5\sqrt{5} = 2500\sqrt{5} \, \text{cm}^3 V5590.17cm3V \approx 5590.17 \, \text{cm}^3


Final Answer:

The largest possible volume of the box is approximately 5590.17 cm³.


Would you like detailed explanations of any specific step?

Related Questions:

  1. What would happen to the volume if the material were doubled?
  2. How does the height of the box change if the base size changes?
  3. What is the relationship between surface area and volume for this shape?
  4. How do derivatives help in finding maximum values in such problems?
  5. Could the box have a closed top with the same material constraint?

Tip:

Always check the constraints carefully when solving optimization problems, as they dictate feasible solutions.

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Math Problem Analysis

Mathematical Concepts

Optimization
Derivatives
Surface Area
Volume of 3D Shapes

Formulas

Surface area constraint: x^2 + 4xh = 1500
Volume of the box: V = x^2h
Optimization condition: dV/dx = 0

Theorems

Extreme Value Theorem
First Derivative Test

Suitable Grade Level

Grades 11-12 (High School Calculus)