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[SOLVED] maximum problem

DeusAbscondus

Active member
Jun 30, 2012
176
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
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Take another look at the linear or second term of your derivative...
 

Amer

Active member
Mar 1, 2012
275
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
[tex] V' = 12x^2 - 120 x + 200 [/tex]
[tex] 12x^2 - 120x + 200 = 0 [/tex]
[tex] 3x^2 - 30 + 50 = 0 [/tex]
[tex] x = \frac{30 \mp \sqrt{900 - 4(3)(50}}{6} = \frac{ 30 \mp 10 \sqrt{3}}{6} [/tex]
study the sign of V' around it is zeros we will get it has a maximum at [tex] x = \frac{30 - 10\sqrt{3}}{6} [/tex]
 

DeusAbscondus

Active member
Jun 30, 2012
176
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)