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A memory chip is being designed to hold a number of transistors and heat sinks. The transistors hold memory while the heat sinks cool the chip. There must be at least one heat sink for every 2,000 transistors to prevent overheating. Also, each transistor has an area of 
2.0 ×10^(-10)mm^(2) (square millimeters), each heat sink has an area of 
3.6 ×10^(-6)mm^(2), and the total area of transistors and heat sinks must be at most 
2mm^(2). What is the approximate maximum number of transistors that the chip can hold according to this design?
Choose 1 answer:
(A) 
2.78 ×10^(2)
(B) 
5.56 ×10^(5)
(C) 
1.0 ×10^(9)
(D) 
1.0 ×10^(10)

A memory chip is being designed to hold a number of transistors and heat sinks. The transistors hold memory while the heat sinks cool the chip. There must be at least one heat sink for every 2,0002,000 transistors to prevent overheating. Also, each transistor has an area of 2.0×1010mm22.0 \times 10^{-10}\text{mm}^{2} (square millimeters), each heat sink has an area of 3.6×106mm23.6 \times 10^{-6}\text{mm}^{2}, and the total area of transistors and heat sinks must be at most 2mm22\text{mm}^{2}. What is the approximate maximum number of transistors that the chip can hold according to this design?\newline Choose 11 answer:\newline (A) 2.78×1022.78 \times 10^{2}\newline (B) 5.56×1055.56 \times 10^{5}\newline (C) 1.0×1091.0 \times 10^{9}\newline (D) 1.0×10101.0 \times 10^{10}

Full solution

Q. A memory chip is being designed to hold a number of transistors and heat sinks. The transistors hold memory while the heat sinks cool the chip. There must be at least one heat sink for every 2,0002,000 transistors to prevent overheating. Also, each transistor has an area of 2.0×1010mm22.0 \times 10^{-10}\text{mm}^{2} (square millimeters), each heat sink has an area of 3.6×106mm23.6 \times 10^{-6}\text{mm}^{2}, and the total area of transistors and heat sinks must be at most 2mm22\text{mm}^{2}. What is the approximate maximum number of transistors that the chip can hold according to this design?\newline Choose 11 answer:\newline (A) 2.78×1022.78 \times 10^{2}\newline (B) 5.56×1055.56 \times 10^{5}\newline (C) 1.0×1091.0 \times 10^{9}\newline (D) 1.0×10101.0 \times 10^{10}
  1. Constraints: Understand the constraints for the memory chip design.\newline- There must be at least one heat sink for every 2,0002,000 transistors.\newline- Each transistor has an area of 2.0×1010 mm22.0 \times 10^{-10} \text{ mm}^2.\newline- Each heat sink has an area of 3.6×106 mm23.6 \times 10^{-6} \text{ mm}^2.\newline- The total area for transistors and heat sinks must be at most 2 mm22 \text{ mm}^2.
  2. Equations Setup: Let TT be the number of transistors and HH be the number of heat sinks. According to the constraints, we have:\newlineHT2000H \geq \frac{T}{2000}
  3. Total Area Inequality: Express the total area in terms of TT and HH and set up the inequality based on the maximum area allowed.\newlineTotal area = (Area of one transistor ×\times Number of transistors) + (Area of one heat sink ×\times Number of heat sinks)\newlineTotal area = (2.0×1010 mm2×T)+(3.6×106 mm2×H)(2.0 \times 10^{-10} \text{ mm}^2 \times T) + (3.6 \times 10^{-6} \text{ mm}^2 \times H)\newlineTotal area 2 mm2\leq 2 \text{ mm}^2
  4. Substitution: Substitute the relationship between TT and HH into the total area inequality.\newlineSince HT/2000H \geq T / 2000, we can use H=T/2000H = T / 2000 in the inequality:\newline(2.0×1010 mm2×T)+(3.6×106 mm2×(T/2000))2 mm2(2.0 \times 10^{-10} \text{ mm}^2 \times T) + (3.6 \times 10^{-6} \text{ mm}^2 \times (T / 2000)) \leq 2 \text{ mm}^2
  5. Solving for T: Simplify the inequality to solve for T.\newline(2.0×1010 mm2×T)+(3.6×106 mm2×T/2000)2 mm2(2.0 \times 10^{-10} \text{ mm}^2 \times T) + (3.6 \times 10^{-6} \text{ mm}^2 \times T / 2000) \leq 2 \text{ mm}^2\newline(2.0×1010T)+(1.8×109T)2(2.0 \times 10^{-10} T) + (1.8 \times 10^{-9} T) \leq 2\newline(2.0×1010+1.8×109)T2(2.0 \times 10^{-10} + 1.8 \times 10^{-9}) T \leq 2\newline(2×1010+18×1010)T2(2 \times 10^{-10} + 18 \times 10^{-10}) T \leq 2\newline(20×1010)T2(20 \times 10^{-10}) T \leq 2\newlineT2/(20×1010)T \leq 2 / (20 \times 10^{-10})\newlineT2/(2×109)T \leq 2 / (2 \times 10^{-9})\newlineT1×109T \leq 1 \times 10^9
  6. Final Answer: Choose the answer that is closest to the maximum number of transistors without exceeding it.\newlineThe closest answer that does not exceed 1×1091 \times 10^9 is:\newline(C) 1.0×1091.0 \times 10^9

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