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 2.0×10−10mm2 (square millimeters), each heat sink has an area of 3.6×10−6mm2, and the total area of transistors and heat sinks must be at most 2mm2. What is the approximate maximum number of transistors that the chip can hold according to this design? Choose 1 answer: (A) 2.78×102 (B) 5.56×105 (C) 1.0×109 (D) 1.0×1010
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 2,000 transistors to prevent overheating. Also, each transistor has an area of 2.0×10−10mm2 (square millimeters), each heat sink has an area of 3.6×10−6mm2, and the total area of transistors and heat sinks must be at most 2mm2. What is the approximate maximum number of transistors that the chip can hold according to this design? Choose 1 answer: (A) 2.78×102 (B) 5.56×105 (C) 1.0×109 (D) 1.0×1010
Constraints: Understand the constraints for the memory chip design.- There must be at least one heat sink for every 2,000 transistors.- Each transistor has an area of 2.0×10−10 mm2.- Each heat sink has an area of 3.6×10−6 mm2.- The total area for transistors and heat sinks must be at most 2 mm2.
Equations Setup: Let T be the number of transistors and H be the number of heat sinks. According to the constraints, we have:H≥2000T
Total Area Inequality: Express the total area in terms of T and H and set up the inequality based on the maximum area allowed.Total area = (Area of one transistor × Number of transistors) + (Area of one heat sink × Number of heat sinks)Total area = (2.0×10−10 mm2×T)+(3.6×10−6 mm2×H)Total area ≤2 mm2
Substitution: Substitute the relationship between T and H into the total area inequality.Since H≥T/2000, we can use H=T/2000 in the inequality:(2.0×10−10 mm2×T)+(3.6×10−6 mm2×(T/2000))≤2 mm2
Solving for T: Simplify the inequality to solve for T.(2.0×10−10 mm2×T)+(3.6×10−6 mm2×T/2000)≤2 mm2(2.0×10−10T)+(1.8×10−9T)≤2(2.0×10−10+1.8×10−9)T≤2(2×10−10+18×10−10)T≤2(20×10−10)T≤2T≤2/(20×10−10)T≤2/(2×10−9)T≤1×109
Final Answer: Choose the answer that is closest to the maximum number of transistors without exceeding it.The closest answer that does not exceed 1×109 is:(C) 1.0×109
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