Optimization ( Applied Max and Minimum )

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SUMMARY

The discussion focuses on optimizing the volume of an open-topped box constructed from a square piece of cardboard measuring 30 cm on each side. By cutting out squares of side length x from each corner, the volume function is defined as V(x) = x(30 - 2x)². The critical points are determined through calculus, yielding dimensions of 5 cm for height and 20 cm for both length and width, which maximize the box's volume. The second derivative test confirms that this configuration provides a maximum volume.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and critical points.
  • Familiarity with volume optimization problems involving geometric shapes.
  • Ability to manipulate polynomial equations and perform second derivative tests.
  • Knowledge of the implications of boundary conditions in optimization.
NEXT STEPS
  • Study the application of the second derivative test in optimization problems.
  • Explore volume optimization techniques for different geometric shapes.
  • Learn about the implications of boundary conditions in calculus-based optimization.
  • Investigate real-world applications of optimization in manufacturing and design.
USEFUL FOR

Students studying calculus, educators teaching optimization techniques, and anyone interested in practical applications of mathematical principles in design and engineering.

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Homework Statement



From a square piece of cardboard, 30 cm on each side, an open topped box is to be constructed by cutting the squares from the corners and turning up the sides. What are the dimensions of the box of largest volume?





The Attempt at a Solution



I know how to do the derivatives from the equation below. but from the book I don't get why the base of the box is : 30 - 2x.

The equation goes like this:
V(x) = x(30-2x)^2.
 
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base of the box would be 30 - 2x because you are subtracting x twice. Try to draw a diagram to help you

so like you said

v(x) = x(30-2x)(30-2x)
v(x) = 4x^3 - 120x^2 + 900x
v`(x) = 12x^2 - 240x + 900
v`(x) = 12(x-15)(x-5)
v`(x) = 5, 15

Therefore dimestions of largest volume are 5, 20, 20

[note the interval was 0 < x < 15 (if you went higher than 15 you would have negative distance, you can close the intervals or open them depending on how you look at it- but in this case 15 would yield a minimum rather than a maximum)

Check:
v``(x) = 24x - 240
v``(5) = 24(5) < 240
so this is a maximum because v``(x) < 0
 

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