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

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SUMMARY

The discussion centers on finding the optimum value for \(x\) in the volume equation of a rectangular prism, \(V=4x^3-60x^2+200x\). The derivative, \(V'=12x^2-120x+200\), is set to zero to locate the turning point, leading to the quadratic equation \(3x^2-30x+50=0\). The correct solution for \(x\) is approximately \(2.11\), yielding a maximum volume of approximately \(192.45 \, \text{cm}^3\). Participants emphasized the importance of correctly applying the quadratic formula and factoring techniques to arrive at the solution.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and critical points.
  • Familiarity with quadratic equations and the quadratic formula.
  • Basic knowledge of volume calculations for geometric shapes.
  • Ability to graph functions to visualize maxima and minima.
NEXT STEPS
  • Study the application of the quadratic formula in solving polynomial equations.
  • Learn about optimization techniques in calculus, particularly for cubic functions.
  • Explore graphical methods for finding maxima and minima of functions.
  • Review the concepts of critical points and their significance in determining function behavior.
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Students studying calculus, educators teaching optimization problems, and anyone interested in applying mathematical concepts to real-world volume calculations.

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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