Answer:
Solution
verified
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
2
=
4
3
x
2
+x(
2
16−3x
)
A=8x+(
4
3
−
2
3
)x
2
dx
dA
=8+(
4
3
−
2
3
)2x
For maxima or minima,
dx
dA
=0
⇒4−
4
(6−
3
)
x=0.
∴x=
6−
(3)
16
=
36−3
16(6+
3
)
=
33
16(6+1.73)
=
33
16(7.73)
=
33
123.68
⇒x=3.75 nearly.
Now,
dx
2
d
2
A
=2(
4
3
−
2
3
)<0
Hence A is maximum.
By (1),
y=2.375
Step-by-step explanation:
