The line segment length squared is a^2 + b^2 = 20^2 = 400.
The area of the triangle is (1/2)a b
A^2 = (1/4)a^2 b^2 = (1/4)a^2(400 - a^2)
= 100 a^2 - a^4]/4
When a maximum area (or A^2) is achieved,
d(A^2)/da = 0 = 200 a- a^3
The other solution, a =0, yields a minumum area (0), not a maximum.
a^2 = 200, therefore
b^2 = 200
Therefore a = b.
You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a,0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a=b.
Can anyone please give me some ideas to do it????THANKS A LOT!!
2 answers
Tq ,tqs a lot sir/mam
For free solution 😊👍
For free solution 😊👍