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Tutor profile: Mehmet D.

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Mehmet D.
PhD student at UW-Madison, was a math tutor for 1.5 years in college
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Questions

Subject: Discrete Math

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Question:

Assume that $$m$$ is an integer. Prove that $$m\cdot (m+3)$$ is always an even integer.

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Mehmet D.
Answer:

First, assume $$m$$ is an even integer. That is, by definition, $$m=2k$$ for some integer $$k$$. Then, $$m+3 = 2k+3$$. This suggests that $$m(m+3) = (2k)(2k+3) = 2k(2k+3)$$. As $$k$$ is an integer, $$2k+3$$ must be an integer, which means $$k(2k+3)$$ is an integer. Therefore, $$m(m+3) = 2k(2k+3)$$ is even by definition. Now, assume $$m$$ is an odd integer. That is, by definition, $$m=2k+1$$ for some integer $$k$$. Then, $$m+3 = 2k+4$$. This suggests that $$m(m+3) = (2k+1)(2k+4) = 2(2k+1)(k+2)$$. As $$k$$ is an integer, $$2k+1$$ and $$k+2$$ must both be integers, which means $$(2k+1)(k+2)$$ is an integer. Therefore, $$m(m+3) = 2(2k+1)(k+2)$$ is even by definition. As $$m$$ has to be either even or odd, we conclude that $$m(m+3)$$ must be even for all integers $$m$$.

Subject: Computer Science (General)

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Question:

Computers use the binary system, meaning that everything in a computer's memory is stored in bits that consist of $$0$$s and $$1$$s. If we want to be able to store integer values ranging from $$0$$ to $$255$$, how many binary bits do we need at the very least?

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Mehmet D.
Answer:

We need 8 bits to be able to represent numbers from $$0$$ to $$255$$. This is because the biggest binary number that can be written in $$8$$ bits is $$11111111$$, which corresponds to $$1\times2 ^0 + 1\times2 ^1 + 1\times2 ^2 + 1\times2 ^3 + 1\times2 ^4 + 1\times2 ^5 + 1\times2 ^6 + 1\times2 ^7 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$$. The smallest $$8$$-bit number is $$0000000$$, which corresponds to 0. Therefore, an $$8$$-bit number is sufficient to represent all integer values from $$0$$ to $$255$$.

Subject: Calculus

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Question:

A rectangle with a perimeter of $$36$$ cm has integer dimensions, i.e., its width and height are integer-valued. a) What should be the width and height of this rectangle if we want to maximize its area? b) What if the sides of this rectangle cannot be the same? Answer this part without involving derivatives.

Inactive
Mehmet D.
Answer:

a) Let's denote the sides of this rectangle by $$w$$ and $$h$$, which we assume to be integer-valued according to the question. If the perimeter of this rectangle is $$36$$ cm, then $$2\cdot(w + h) = 36$$ must be true, which means $$w + h = 18$$. To simplify things down to one variable, let's write $$h$$ in terms of $$w$$ based on this equation, i.e., $$h = 18 - w$$. Now, the area of a rectangle is $$A = (\text{width} \times \text{height})$$. Then, for our rectangle, we have $$A = w \cdot (18 - w) = 18w-w^2$$. To find $$w$$ that maximizes $$A$$, we have to take its derivative and set it to zero. By the power rule, we have $$\frac{\partial A}{\partial w} = 18 - 2w$$. Setting it equal to $$0$$ will gives us $$18-2w=0 \implies w=9$$. Since we know that $$h = 18 - w$$, we find that $$h=9$$. Therefore, $$w=9$$ cm and $$h=9$$ cm are the dimensions of the rectangle with a perimeter of $$36$$ cm that has the largest possible area ($$A = 9 \cdot 9 = 81$$). Let's see if this makes sense intuitively. Say we have this rectangle, or rather square, with the dimensions $$w=9$$ cm and $$h=9$$ cm. Now, imagine we can play with this rectangle's dimensions while keeping its perimeter fixed at $$36$$ cm. In other words, if we make its width smaller, we have to make its height bigger. Then, if we keep doing this, we are going to approach to a vertical line with zero area. If we keep increasing its width while decreasing its height, we will approach to a horizontal line with zero area. So, both of these extreme endpoints give us an area of $$0$$. Therefore, it makes sense to have a rectangle with equal sides in order to make the area size as large as possible. b) Now, we imagine that $$w$$ and $$h$$ cannot be equal to each other. However, we still want to maximize the area size while keeping the perimeter at $$36$$ cm. Then, given that the sides have to be integer-valued, we can decrease either one of the optimal $$w$$ or $$h$$ values by $$1$$ while increasing the other one by $$1$$, e.g., $$w=10$$ cm and $$h=8$$ cm ($$A = 10 \cdot 8 = 80$$).

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