5) A rectangle has a perimeter of 60 m. What could be the maximum area of the rectangle?
Hint: use
[tex]k = \frac{4ac - {b}^{2} }{4a} [/tex]
Hint: use
[tex]k = \frac{4ac - {b}^{2} }{4a} [/tex]
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Answer:
225
Step-by-step explanation:
L using 60=2L+2W. Plugging in W=15 we get 60=2L+2(15) which simplifies to L=15. Thus, the area A=15*15=225.
I'm not sure
Hope it's help
Solution:
l=P
l=P2﹣w