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I want to take derivative with respect to $p(t)$, but I am not sure if I can just assume $p(t)$ is another variable since it depends on $t$.

$ \pi = \int_a^b p(t)\cdot \bigl(a-b\cdot p(t)\bigr)\cdot(u- v \cdot t)\, dt $ Thanks

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    I am sorry, I am not clear what to do. I don't know the function form of p(t). What I am trying to do is to take derivative with respect to$p(t)$and equate that with zero to find the optimal function form for p(t). Would this clarification make any change in your response? Thanks2012-05-16

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Okay, I'll expand my answer. For any fixed functions $p$ and $\phi$ the expression $\int_a^b (p(t)+h\phi(t))(a−bp(t)-bh\phi(t))(u−vt)dt$ is a function of the real variable $h$. So we can take derivative with respect to $h$ and equate that to $0$. If you are unsure about legitimacy of taking $\frac{d}{dh}$ under the integral sign, just expand the product and move the powers of $h$ out of the integrals. Like this: $\int_a^b p(t)(a−bp(t))(u−vt)dt + h\left(\int_a^b \phi(t)(a−bp(t))(u−vt)dt + \int_a^b p(t)(-b\phi(t))(u−vt)dt\right) + h^2 \int_a^b \phi(t)(-b\phi(t))(u−vt)dt $ If $p$ is an extremal function for this functional, the derivative $\frac{d}{dh}$ will be zero when $h=0$. So, $\int_a^b \phi(t)(a−bp(t))(u−vt)dt + \int_a^b p(t)(-b\phi(t))(u−vt)dt = 0$ (You notice that the effort put into extracting $h^2$ was wasted.) Combine the integrals and factor out $\phi$: $\int_a^b \phi(t)\left[(a−bp(t))(u−vt)-bp(t)(u-vt)\right]\,dt = 0$ Since $\phi$ could be any integrable function, the expression in square brackets must be $0$ identically. This gives you an equation for $p$.

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    @Elnaz It's still a function; a *constant* function. The answer is not surprising since you imposed no constraints on $p$. If the function $p$ is free to attain any values it pleases, it will simply maximize the product $p(t)(a-b p(t))$ for each $t$; the maximum is attained when $p(t)=a/(2b)$. So, you should consider whether there are any constraints on $p$. Also think about the sign of $u-vt$, and whether you want to maximize or minimize $\pi$.2012-05-17