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

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The discussion focuses on finding the optimum value of \( x \) for 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 identify turning points. Participants discuss using the quadratic formula and factoring to solve for \( x \), ultimately confirming that the maximum volume occurs at approximately \( x = 2.11 \) with a volume of about \( 192.45 \, cm^2 \). Clarifications on the derivative's terms and the correct application of the quadratic formula are provided. The thread concludes with a resolution to the initial confusion regarding the 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)