Find upper bound for $frac{b(abc)^2 (1+bx)x}{a((1+s)(1+bx)+bs(T-x))^2}$
I want to find an expression for an upper bound for $$F(x) = frac{b(abs)^2 (1+bx)x}{a((1+s)(1+bx)+bs(T-x))^2},$$
where $a,b,s, T$ are constants, $a,b,s$ are positive, and $x leq T$, and subject to the constraint that $frac{abs}{1+s(1+bx)} leq K$.
Here's what I did:
The values of the numerator varies as $x(bx+1)$ varies, which is continuous everywhere and positive for positive values of $x$. Hence, the value of the numerator is maximum when $x=T$. This also makes $bs(T-x)$ in the denominator vanish. So we have
begin{align*}
F(x) & leq frac{b(abs)^2 (1+bT)T}{a((1+s)(1+bT))^2}\
& = frac{b(abs)^2 T}{a(1+s)^2(1+bT)}\
& leq frac{bK^2 (1+s(1+bT))^2 T}{a(1+s)^2(1+bT)}
end{align*}
The last inequality comes from the constraint involving $K$.
Now, I realize that my reasoning hinges on the fact that $x$ is positive. How can I improve this solution of mine?
upper-lower-bounds
asked Nov 19 '18 at 4:28
user161300
666
add a comment |
I want to find an expression for an upper bound for $$F(x) = frac{b(abs)^2 (1+bx)x}{a((1+s)(1+bx)+bs(T-x))^2},$$
where $a,b,s, T$ are constants, $a,b,s$ are positive, and $x leq T$, and subject to the constraint that $frac{abs}{1+s(1+bx)} leq K$.
Here's what I did:
The values of the numerator varies as $x(bx+1)$ varies, which is continuous everywhere and positive for positive values of $x$. Hence, the value of the numerator is maximum when $x=T$. This also makes $bs(T-x)$ in the denominator vanish. So we have
begin{align*}
F(x) & leq frac{b(abs)^2 (1+bT)T}{a((1+s)(1+bT))^2}\
& = frac{b(abs)^2 T}{a(1+s)^2(1+bT)}\
& leq frac{bK^2 (1+s(1+bT))^2 T}{a(1+s)^2(1+bT)}
end{align*}
The last inequality comes from the constraint involving $K$.
Now, I realize that my reasoning hinges on the fact that $x$ is positive. How can I improve this solution of mine?
upper-lower-bounds
asked Nov 19 '18 at 4:28
user161300
666
add a comment |
I want to find an expression for an upper bound for $$F(x) = frac{b(abs)^2 (1+bx)x}{a((1+s)(1+bx)+bs(T-x))^2},$$
where $a,b,s, T$ are constants, $a,b,s$ are positive, and $x leq T$, and subject to the constraint that $frac{abs}{1+s(1+bx)} leq K$.
Here's what I did:
The values of the numerator varies as $x(bx+1)$ varies, which is continuous everywhere and positive for positive values of $x$. Hence, the value of the numerator is maximum when $x=T$. This also makes $bs(T-x)$ in the denominator vanish. So we have
begin{align*}
F(x) & leq frac{b(abs)^2 (1+bT)T}{a((1+s)(1+bT))^2}\
& = frac{b(abs)^2 T}{a(1+s)^2(1+bT)}\
& leq frac{bK^2 (1+s(1+bT))^2 T}{a(1+s)^2(1+bT)}
end{align*}
The last inequality comes from the constraint involving $K$.
Now, I realize that my reasoning hinges on the fact that $x$ is positive. How can I improve this solution of mine?
upper-lower-bounds
asked Nov 19 '18 at 4:28
user161300
666
I want to find an expression for an upper bound for $$F(x) = frac{b(abs)^2 (1+bx)x}{a((1+s)(1+bx)+bs(T-x))^2},$$
where $a,b,s, T$ are constants, $a,b,s$ are positive, and $x leq T$, and subject to the constraint that $frac{abs}{1+s(1+bx)} leq K$.
Here's what I did:
The values of the numerator varies as $x(bx+1)$ varies, which is continuous everywhere and positive for positive values of $x$. Hence, the value of the numerator is maximum when $x=T$. This also makes $bs(T-x)$ in the denominator vanish. So we have
begin{align*}
F(x) & leq frac{b(abs)^2 (1+bT)T}{a((1+s)(1+bT))^2}\
& = frac{b(abs)^2 T}{a(1+s)^2(1+bT)}\
& leq frac{bK^2 (1+s(1+bT))^2 T}{a(1+s)^2(1+bT)}
end{align*}
The last inequality comes from the constraint involving $K$.
Now, I realize that my reasoning hinges on the fact that $x$ is positive. How can I improve this solution of mine?
upper-lower-bounds
upper-lower-bounds
asked Nov 19 '18 at 4:28
user161300
666
asked Nov 19 '18 at 4:28
user161300
666
asked Nov 19 '18 at 4:28
user161300
666
asked Nov 19 '18 at 4:28
user161300
666
asked Nov 19 '18 at 4:28
user161300
666
666
add a comment |
add a comment |
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