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secondary 4 | A Maths
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Anyone knows how to do? Can you help out please
Date Posted: 4 years ago
Views: 251
It is interesting to know that the total length of wire required to wrap around the perimeter of the rectangle is a minimum when the rectangle is a square (i.e. the length equals the breadth), regardless of the area of the rectangle given.
If the rectangle measures x cm by 144/x cm, then the total perimeter is 2x + 288/x2 cm.
We can simply perform differentiation (like Thomas has done) or we can plot the graph of y = 2x + 288/x2 on a sheet of graph paper.
If the rectangle measures x cm by 144/x cm, then the total perimeter is 2x + 288/x2 cm.
We can simply perform differentiation (like Thomas has done) or we can plot the graph of y = 2x + 288/x2 on a sheet of graph paper.
Yeah it kind of fazed me when I realised that the length=breadth. I had a misconception and thought the question was asking for the length (l) of the rectangle, not the perimeter (p)
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hey, thank you for helping out! However, id like to clarify why you subbed 12 into P instead of 144/x?
Because they are trying to minimise perimeter. So u sub the stat value of x into the thing u are trying to max or min
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