Answer:
Solution
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Verified by Toppr
Correct option is C)
Perimeter of window P=2y+3x=16
⇒y=
2
16−3x
....(1)
Area A=xy+
4
3
x
=
+x(
)
A=8x+(
−
)x
dx
dA
=8+(
)2x
For maxima or minima,
=0
⇒4−
(6−
x=0.
∴x=
6−
(3)
16
36−3
16(6+
33
16(6+1.73)
16(7.73)
123.68
⇒x=3.75 nearly.
Now,
d
A
=2(
)<0
Hence A is maximum.
By (1),
y=2.375
Explanation:
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