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Yuriy S.

theoretical scientist

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Linear Programming

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Question:

\textit{\textbf{Transportation scheduling}}. Your publishing company is about to start a promotional blitz for its new book. You have 20 salespeople stationed in Chicago and 10 in Denver. You would like to fly at most 10 salespeople into Los Angeles and at most 15 into New York. A round-trip plane flight from Chicago to Los Angeles costs \$195; one from Chicago to New York costs \$182; one from Denver to Los Angeles costs \$395; and one from Denver to New York costs \$166. You want to spend at most \$4,520 on plane flights. How many salespeople should you fly from each of Chicago and Denver to each of Los Angeles and New York to have the most salespeople on the road?

Yuriy S.

Answer:

\textit{\textbf{Setting up a linear programming problem}}. \textit{Find}: the numbers of people flying, $x_{1}$ – from Chicago to Los Angeles, $x_{2}$ – from Chicago to New York, $x_{3}$ – from Denver to Los Angeles, $x_{4}$ – from Denver to New York. \textit{Maximize}: total number of people arriving to both Los Angeles and New York, $x_{1} + x_{2} + x_{3} + x_{4}$. \textit{Constraints}: 1) No more than 20 salespeople flying from Chicago: $x_{1} + x_{2} \leq 20$. 2) No more than 10 salespeople flying from Denver: $x_{3} + x_{4} \leq 10$. 3) At most 10 salespeople flying to Los Angeles: $x_{1} + x_{3} \leq 10$. 4) At most 15 to New York: $x_{2} + x_{4} \leq 15$. 5) Spending at most \$4,520 on flights: $195x_{1} + 182 x_{2} + 395 x_{3} + 166 x_{4} \leq 4520$. 6) $x_{1} \geq 0$. 7) $x_{2} \geq 0$. 8) $x_{3} \geq 0$. 9) $x_{4} \geq 0$. \textit{\textbf{Feasible region}}. Solving all constraints apart from 5), we obtain $0 \leq x_{1} \leq 10, 0 \leq x_{2} \leq 20 – x_{1}, 0 \leq x_{3} \leq 10 – x_{1}, 0 \leq x_{4} \leq 10 – x_{3}$. From constraint 5), we obtain additional condition $x_{4} \leq \frac{2260}{83} – \frac{195}{166} x_{1} – \frac{91}{83} x_{2} – \frac{395}{166} x_{3}$. \textit{\textbf{Solution}}. From the feasible region, we obtain $0 \leq x_{1} + x_{2} \leq 20, 0 \leq x_{3} + x_{4} \leq 10$. Maximization of $x_{1} + x_{2} + x_{3} + x_{4}$ can be done, e.g., by testing different values of $x_{1} + x_{2}$ and $x_{3} + x_{4}$, starting from the maximal values of these sums. For example, substituting $x_{1} + x_{2} = 20$ and $x_{3} + x_{4} = 10$ into constraint 5), we find that $13 x_{1} + 229 x_{3} \leq -780$, which is impossible for non-negative $x_{1}$ and $x_{2}$. Next we can continue in ascending order of $x_{1} + x_{2} + x_{3} + x_{4}$ by testing the following two cases: ${x_{1} + x_{2} = 19, x_{3} + x_{4} = 10}$ and ${x_{1} + x_{2} = 20, x_{3} + x_{4} = 9}$. Maximum number of people flying is 25: $x_{1} = 10, x_{2} = 5, x_{3} = 0, x_{4} = 10$. The amount spent is \$4,520.

Calculus

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Question:

Apply change of variable to find $\int{(x^{4}-7)x^{3}dx}$.

Yuriy S.

Answer:

Let $u=x^{4}-7$. Then $du = 4x^{3}dx$, hence $x^{3}dx = du/4$. The integral simplifies to $\frac{1}{4}\int{udu} = \frac{u^{2}}{8}+C = \frac{(x^{4}-7)^{2}}{8}+C$, where $C$ is an arbitrary integration constant.

Algebra

TutorMe

Question:

Let $f(x) = x^{3}-6x+4$, $g(x) = 7x+1$. Find the function $f \cdot g$, its domain and range.

Yuriy S.

Answer:

This is a composition of two functions, $f$ and $g$. First, the function $g$ on the right side of the composition is applied to the argument $x$, and then the next function to the left of it, i.e., $g$ is applied to the result. Thus, $f \cdot g$ means $f(g(x))$. Substituting the definition of $g(x)$, one obtains $f(7x+1)$. Further, to apply $f(x)$, its definition is used, where the argument $x$ is simply replaced with the required argument $(7x+1)$, resulting in $f \cdot g (x) = (7x+1)^{3}-6(7x+1)+4 = 343x^{3}+147x^{2}-21x-1$. The result is a polynomial, so it has no singular points or ranges of the argument $x$, which should be excluded from its domain. Thus, the domain is $(-\infty, \infty)$ -- the set of all real numbers. Since the result is a cubic polynomial, it can take all values from $-\infty$ to $\infty$, so its range is $(-\infty, \infty)$.

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