Maximizing Volume of a Rectangular Prism: Can't Get Optimum Value for $x$

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Discussion Overview

The discussion revolves around finding the optimum value for \(x\) in the volume equation of a rectangular prism, given by \(V=4x^3-60x^2+200x\). Participants explore methods to derive the critical points of the volume function through calculus, particularly using the derivative \(V'\) and setting it to zero to identify turning points. The conversation includes attempts to apply the quadratic formula and graphing techniques.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in obtaining the optimum value for \(x\) using the quadratic formula, despite having the correct derivative \(V'=12x^2-60x+200\).
  • Another participant points out an inconsistency in the linear term of the derivative, suggesting a need for reevaluation.
  • A later reply provides a corrected form of the derivative as \(V' = 12x^2 - 120x + 200\) and attempts to solve it using the quadratic formula, leading to a proposed solution of \(x = \frac{30 \mp 10\sqrt{3}}{6}\).
  • One participant acknowledges their earlier confusion and confirms they have found the solution, arriving at \(x \approx 2.11\) through factoring and applying the quadratic formula correctly.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to solving for \(x\), as there are multiple interpretations and methods presented. Some participants correct earlier claims while others express ongoing confusion.

Contextual Notes

There are unresolved issues regarding the application of the quadratic formula and the correct formulation of the derivative. Participants have differing views on the steps taken to arrive at the solution.

DeusAbscondus
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No matter how or what strategy I try, I can't get the optimum value for $x$ in the following equation:

$$V=4x^3-60x^2+200x$$ Let V=Volume of a rectangular prism, so:
$$V'=12x^2-60x+200$$ Set V'=0 to get turning point
$$12x^2-60x+200=0$$
The answer given in my text is when
$$ x=2.11 \Rightarrow \text{ the maximum volume is }\approx 192.45cm^2$$

I can graph it to get this, but can't get it using the quadratic formula for some reason.

some help would be appreciated
 
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DeusAbscondus said:
No matter how or what strategy I try, I can't get the optimum value for $x$ in the following equation:

$$V=4x^3-60x^2+200x$$ Let V=Volume of a rectangular prism, so:
$$V'=12x^2-60x+200$$

\[ V'=12x^2-120x+200 \]

CB
 
Take another look at the linear or second term of your derivative...
 
DeusAbscondus said:
No matter how or what strategy I try, I can't get the optimum value for $x$ in the following equation:

$$V=4x^3-60x^2+200x$$ Let V=Volume of a rectangular prism, so:
$$V'=12x^2-60x+200$$ Set V'=0 to get turning point
$$12x^2-60x+200=0$$
The answer given in my text is when
$$ x=2.11 \Rightarrow \text{ the maximum volume is }\approx 192.45cm^2$$

I can graph it to get this, but can't get it using the quadratic formula for some reason.

some help would be appreciated

V' = 12x^2 - 120 x + 200
12x^2 - 120x + 200 = 0
3x^2 - 30 + 50 = 0
x = \frac{30 \mp \sqrt{900 - 4(3)(50}}{6} = \frac{ 30 \mp 10 \sqrt{3}}{6}
study the sign of V' around it is zeros we will get it has a maximum at x = \frac{30 - 10\sqrt{3}}{6}
 
DeusAbscondus said:
No matter how or what strategy I try, I can't get the optimum value for $x$ in the following equation:Thanks people; what a goose am I; had been staring at this way too long; now I've got it:
$$V=4x^3-60x^2+200x$$ Let V=Volume of a rectangular prism, so:
$$V'=12x^2-120x+200$$ Set V'=0 to get turning point
$$12x^2-120x+200=0$$ (simply by factoring by 4 then apply quadratic formula)
$$3x^2-30x+50=0$$
$$\Rightarrow \frac{30\pm\sqrt{900-600}}{6}$$
$$=\frac{30\pm\sqrt{300}}{6}\Rightarrow \frac{30-\sqrt{300}}{6}\approx2.11$$(Dance)(Rofl)
 

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