Finding Extreme Volumes of V(x): A Box Model

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SUMMARY

The discussion focuses on finding the extreme values of the volume function V(x) = x(14-2x)(15-x) for the interval 0 < x < 7. The derivative of the function is calculated as 210 - 88x + 6x², which is set to zero to find critical points. The next step involves applying the quadratic formula to solve for x, allowing for the determination of whether the critical points represent minimum or maximum volumes within the specified range.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and critical points.
  • Familiarity with the quadratic formula for solving polynomial equations.
  • Knowledge of volume modeling in mathematics.
  • Ability to analyze functions within a defined interval.
NEXT STEPS
  • Apply the quadratic formula to solve 0 = 210 - 88x + 6x² for x.
  • Evaluate the volume function V(x) at the critical points and endpoints of the interval.
  • Determine the nature of the critical points (minimum or maximum) using the second derivative test.
  • Explore applications of box volume optimization in real-world scenarios.
USEFUL FOR

Students studying calculus, mathematics educators, and anyone involved in optimization problems related to volume modeling.

Jan Hill
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Homework Statement



The function V(x) = x(14-2x)(15-x) 0<x<7 models the volume of a box
Find the extreme values of V

and tell whether these represent the smallest and largest volumes or only the smallest or only the largest volumes

Homework Equations





The Attempt at a Solution




I found the derivative of the function to be 210 - 88x + 6x^2 and if I set this equal to 0,
I get 0 = 210 -88x + 6x^2
210 = 88x -6x^2

but where do I go from here?
 
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solve for x using quadratic formula.
 

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