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Determine the value of x so that the rectangle has max. area, where A=xy, y= 8-x^2, 0
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A=x(8-x^2) explicit way is to solving \( \large \frac{d}{dx} A=0 \) and then find maximum of A
is possible for you to solve for max of A, if not i understand. i appreciate the help
actually u need to find the abs. max of A on the interval (0,2sqrt2] first step is Finding critical points. Compute A'(x)
okay thanks
A(x)=8x-x^3 and A'(x)=8-3x^2 A'(x)=0 -----> 8=3x^2 ---> \(\large x=\pm \frac{\sqrt{3}}{2\sqrt{2}} \) we choose positive sign because negative sign is not on our interval
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now we must check \( \sqrt{2} \) and \(\large \frac{\sqrt{3}}{2\sqrt{2}} \)
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