The technology company that you work for is designing a new microchip to be used in school classrooms. Your job as the employee is to determine the specs of the chip. They have told you that it will be very thin and rectangular and must have a surface area of 40 mm2 to cut down on the amount of expensive metals used. You’ve also been told that the perimeter of the chip will have rubber on the edges to protect it, and that they have exactly 28 mm of rubber to use.
What are the dimensions of the microchip (the length and width)?
Answers
At the center of attention, mingling easily with old friends, was a man with a graying beard, beach-appropriate longish hair, and a muscular build—someone who looked more like a retired linebacker or professional surfer than the architect of some of the most important microchip designs of the past 30 years. The man was Jim Keller, the person Intel is counting on to revitalize its own struggling chip-design enterprise.
Answer:
Step-by-step explanation:
Let l= length and w = width. We know the perimeter Is 28 mm, and this is a
rectangular chip, so 21 + 2w = 28. This simplifies to I + W = 14.
We also know that the area is 40 mm, so I = 40. From here, you can solve it
in a number of ways. For example, from I + w = 14, we have I = 14 - W. Then
W(14-w) = 40 -> 14w-W2 = 40 -> w2 - 14W + 40 = 0 -> (w-10)(W-4) = 0.
So W=10 or w=4. If you plug these back into the perimeter equation, you see that I = 4 or 10.