check the picture below, recall the base is a square, so each side being equally "x".
[tex]\bf V(x)=x^2\left( \cfrac{300-x^2}{4x} \right)\implies V(x)=x^2\left( \cfrac{75}{x}-\cfrac{x}{4}\right)
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V(x)=75x-\cfrac{x^3}{4}\implies \cfrac{dV}{dx}=75-\cfrac{1}{4}\cdot 3x^2\implies \cfrac{dV}{dx}=75-\cfrac{3x^2}{4}
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\cfrac{dV}{dx}=\cfrac{300-3x^2}{4}\impliedby \textit{now, let's set the derivative to 0}
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0=\cfrac{300-3x^2}{4}\implies 0=300-3x^2\implies 3x^2=300\implies x^2=100
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x=\pm\sqrt{100}\implies x=\pm 10[/tex]
now, if you do a first-derivative test on +10, say check 9.99 and 10.01 for example, you'll notice you'd get + and - value respectively, meaning is a maximum.
