$\left\lceil5n\right\rceil + 30n \geq 250$

Question

Solve $\left\lceil5n\right\rceil + 30n \geq 250$ for $n$ (Show the lowest possible value) $0\leq n \leq 10$

Can someone help me with this? I wrote a script in python and I discovered that the lowest value was $\frac{107}{15}$, but I still don't know how to solve this problem using only maths.

Answer 1

It is confusing to use $n$ for what is apparently not an integer.

Prefer $x$ in this case.

So let $y=5x$ and the equation is $\lceil y\rceil+6y\ge 250$

• If $y$ is an integer then $\lceil y\rceil=y$

The equation becomes $\quad 7y\ge 250\iff y\ge 36\iff x\ge 7.2 $

• If $y$ is not an integer then $y=n+r$ with $n$ integer and $r\in]0,1[$.

The equation becomes $\quad n+1+6(n+r)\ge 250\iff 7n+6r\ge 249\iff \begin{cases} n\ge 35\\ 6r\ge249-7\times 35=4\end{cases}$ or $\begin{cases}n\ge 36\\r\ge 0\end{cases}$.

That is $y\ge 35+\frac 23\iff x\ge 7+\frac 2{15}=\frac{107}{15}\ $ or $y\ge 36$

The minimum value between $7.2$ and $\frac {107}{15}$ being the lattest, then this is the solution for the problem.

Answer 2

Let $n=I+f,0\le f<1$

If $f=0,\lceil n\rceil=I$

If $f>0,$

$$250=5(I+1)+30(I+f)$$

$$\iff35(7-I)/30=f$$

Now $0<f<1$

Answer 3

The left side can only change when $n$ is a multiple of $\frac 1{30}$. The ceiling function can increase the term, which decreases the $n$ required. We can ignore the ceiling function, solve for $n$, then decrease $n$ in increments of $\frac 1{30}$ until we find the answer. So $$5n+30n=250\\n=\frac {250}{35}=\frac{50}7$$ is our upper bound. Now just subtract off $\frac 1{30}$ at a time until $\lceil 5n \rceil+30n$ falls below $250$ This sort of piecewise linear function is more suitable to a bisection search than any other if you are doing basic root finding.