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Let $f(x)=\sqrt{\tan x}$ . Then show that the area bounded by $y=f(x),y=f(c),x=0$ and $x=a,0<c<a<90^\circ$ is minimum when $\displaystyle c=\frac{a}{2}.$

what i try

enclosed area $$A=\int^{a}_{0}\bigg(\sqrt{\tan x}-\sqrt{\tan(c)}\bigg)dx$$

$$A(c)=-\int^{c}_{0}\bigg(\sqrt{\tan x}-\sqrt{\tan(c)}\bigg)dx+\int^{a}_{c}\bigg(\sqrt{\tan x}-\sqrt{\tan(c)}\bigg)dx$$

How do i solve it help me please

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    $\begingroup$ The second integral should be from $c$ to $a$ and should be $\sqrt{\tan c} - \sqrt{\tan x}$. To proceed, use the fact that the minimum occurs when derivative is zero and the fact that derivative is the inverse of integration. $\endgroup$ – cr001 Feb 17 at 9:22
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We have $\sqrt{\tan(x)}$ is an increasing function from $0$ to $90^\circ$.

\begin{align}A(c) &= \int_0^c\sqrt{\tan(c)}-\sqrt{\tan(x)}\, dx +\int_c^a\sqrt{\tan(x)}-\sqrt{\tan(c)}\, dx \\ &= c\sqrt{\tan (c)}-(a-c)\sqrt{\tan(c)}-\int_0^c \sqrt{\tan x}\, dx -\int_a^c \sqrt{\tan x}\, dx\\ &=(2c-a)\sqrt{\tan (c)}-\int_0^c \sqrt{\tan x}\, dx -\int_a^c \sqrt{\tan x}\, dx\\ \end{align}

Using the fundamental theorem of Calculus,

$$A'(c)=2\sqrt{\tan c}+\frac{(2c-a)\sec^2 c}{2\sqrt{\tan c}}-2 \sqrt{\tan{c}}$$

I hope you can complete the rest.

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