Consider a cuboid of sides 2x,4x,and 5x and a | vedclass

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(B) The surface area of the cuboid is $2(2x \cdot 4x + 4x \cdot 5x + 5x \cdot 2x) = 2(8x^2 + 20x^2 + 10x^2) = 2(38x^2) = 76x^2$. The surface area of a

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$19:45$

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(B) The surface area of the cuboid is $2(2x \cdot 4x + 4x \cdot 5x + 5x \cdot 2x) = 2(8x^2 + 20x^2 + 10x^2) = 2(38x^2) = 76x^2$. The surface area of a closed hemisphere is $3\pi r^2$. Given $76x^2 + 3\pi r^2 = k$,so $r^2 = \frac{k - 76x^2}{3\pi}$. The total volume $V = 40x^3 + \frac{2}{3}\pi r^3$. Substituting $r = \left(\frac{k - 76x^2}{3\pi}ight)^{1/2}$,we get $V = 40x^3 + \frac{2}{3}\pi \left(\frac{k - 76x^2}{3\pi}ight)^{3/2}$. Differentiating with respect to $x$ and setting $\frac{dV}{dx} = 0$: $120x^2 + \frac{2}{3}\pi \cdot \frac{3}{2} \left(\frac{k - 76x^2}{3\pi}ight)^{1/2} \cdot \left(\frac{-152x}{3\pi}ight) = 0$. $120x^2 = \frac{152x}{3} \left(\frac{k - 76x^2}{3\pi}ight)^{1/2}$. Since $x 
eq 0$,$120x = \frac{152}{3} \cdot r$. $\frac{x}{r} = \frac{152}{360} = \frac{19}{45}$.

Consider a cuboid of sides $2x$,$4x$,and $5x$ and a closed hemisphere of radius $r$. If the sum of their surface areas is a constant $k$,then the ratio $x:r$,for which the sum of their volumes is maximum,is

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