I need to find the area of the largest rectangle possible under the curve of equation $1 - x^4$ with the base on the x-axis. The answers I've seen from other questions similar to this use calculus, but how would you solve it this using trigonometry?
Finding the largest rectangle under the graph of $y = 1 - x^4$ using trigonometry
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algebra-precalculus
trigonometry
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1You can maximize $A = x - x^5$. Which means find the point where $A$ is largest on $A = x - x^5$. – 2017-01-07
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0Could you explain from where you got $x - x^5$? – 2017-01-07
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1@big_red_bus Since $A=2x(1-x^4)=2(x-x^5)$ – 2017-01-07
1 Answers
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Let $(x,1-x^4)$ is a vertex of the rectangle, where $x>0$.
Thus, by AM-GM $$S=2x(1-x^4)=2\left(x-x^5\right)=2\left(\frac{4}{5\sqrt[4]5}-\left(x^5+\frac{4}{5\sqrt[4]5}-x\right)\right)\leq$$ $$\leq\frac{8}{5\sqrt[4]5}-2\left(5\sqrt[5]{x^5\cdot\left(\frac{1}{5\sqrt[4]5}\right)^4}-x\right)=\frac{8}{5\sqrt[4]5}.$$ The equality occurs for $x=\frac{1}{\sqrt[4]5}$, which says that the answer is $\frac{8}{5\sqrt[4]5}$.