A rectangle has its base on the x-axis and its upper two vertices on the parabola y = 49 − x2. What is the largest possible area (in squared units) of the rectangle?

Explanation:
The parabola is a 'mountain'-type (because the coefficient of
x
2
is negative. Also, it is symmetrical in respect to the
y
-axis, because there is no
x
-term.
We can now simplify the problem to finding a rectangle with vertices at
(
0
,
0
)
,
(
x
,
0
)
,
(
0
,
y
)
and
(
x
,
y
)
and then double the
x
-values.
The area will then be
A
=
x
⋅
y

If we substitute the equation of the parabola for
y
:
A
=
x
⋅
(
49
−
x
2
)
=
49
x
−
x
3

To find the extremes (max of min) we need the derivative and set it to zero:
A
'
=
49
−
3
x
2
=
0
→
x
2
=
49
3
→
x
=
√
49
3
≈
4.04
...

(remember we will have to double that)
Use this in the original function:
y
=
49
−
x
2
=
49
−
49
3
=
98
3
≈
32.67

Answer :
Dimensions will be
8.08
x
32.67

graph{49-x^2 [-65.4, 66.33, -13.54, 52.3]}

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