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 December 11th, 2012, 04:46 PM #1 Senior Member   Joined: Mar 2012 Posts: 111 Thanks: 0 Dimensions of max area and area itself? so far i got: y = 49-x^2 Area = 2x(49-x^2) Area = 98x - 2x^3 6x^2 = -98 x^2 = sqrt(98/6) === 2sqrt(7)/(sqrt(6)
 December 11th, 2012, 06:11 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Dimensions of max area and area itself? You have the correct area function: $A(x)=98x-2x^3$ Differentiating and equating to zero, we find: $A'(x)\,=\,98-6x^2\,=\,2$$49-3x^2$$\,=\,0$ So, our critical number is found from: $49-3x^2=0$ Taking the positive root, we find: $x=\frac{7}{\sqrt{3}}\approx4.04$ and so: $y=49-\frac{49}{3}=\frac{98}{3}\approx32.67$ To ensure we have a maximum, observe that: $A''(x)=-6x$, and since our critical number is positive, the second derivative is negative, showing we do have a maximum. Hence: $A_{\text{max}}=2\cdot\frac{7}{\sqrt{3}}\cdot\frac{ 98}{3}=\frac{1372}{3\sqrt{3}}\approx264.04$
 December 11th, 2012, 06:13 PM #3 Senior Member   Joined: Mar 2012 Posts: 111 Thanks: 0 Re: Dimensions of max area and area itself? thanks mark!

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