The largest rectangular area in the parabola, mathematics is superior!!

2x [(1 36)(x 2) +36 ] = area, 2x is the rectangular length on the axis.

00，-1/36)(x^2) +36>0

So according to 2xy x +y

2x [(1 36)(x 2) +36 ] x +[1 36)(x 2) +36 ].

When x=[(-1 36)(x 2) +36 ], take the equal sign.

First, analyze x=[(-1 36)(x2) +36].

After simplification, there are x 2 + 36 x = 1296

The recipe has (x+18) 2=1296+18 2=1296+324=1620

Solve x+18 = 18 root number 5

x = 18 root numbers 5-18

Because x>0, x=18 root numbers 5-18

So the maximum area = x + [1 36) (x 2) +36 ].

18 root No. 5-18) +1 36) (18 root No. 5-18) 2 + 36 ].

18 (root number 5-1) +1 36) 18 (root number 5-1) 36 ].

18 (root number 5-1) +9 (root number 5-1) 36 ].

18 (root number 5-1) +9) (root number 5-1) (root number 5-1) 2*36*(-9) (root number 5-1) +36

324 (6-2 root number 5) + 81 (56-24 root number 5) -648 (6-2 root number 5) + 1296

1944-648 root number 5 + 4536-1944 root number 5-3888 + 1296 root number 5 + 1296

3888-1296 root No. 5

Because the root number 5 is approximately equal to.

So the above equation is approximately equal to.

I don't know if you've learned the image of the function 3 times.

You solve the equation directly.

1 18)x] x 2 -1296) = 0 and then the image according to the 3rd order function.

The maximum value is determined based on the range of x and the image of the 3rd degree function.

It's definitely not 0 and 36.

This cubic function increases first between 0 and 36, then decreases, and then decreases after 36.

Now all you are asking for is the maximum value of the function x=(0,36).

This method may involve the first derivative, which is the easiest, and I won't do the other way. You said you can't use derivatives, but I can't explain it to you.

First, find the first derivative of f(x)=[(-1 18)x] x 2 -1296), obtain a unary quadratic equation, and find the value of x when the unary quadratic equation is equal to 0, and the value of x within (0,36) is the maximum corresponding x value of f(x).

The answer is 72, you see, right? I'll say yes.

The area formed by the parabolic translation can be solved using the piecewise function method, where the total area of all segments can be obtained by adding the area of all segments according to the area formula of each small segment (rectangle area divided by 2, triangle area).

For example, if a parabola consists of a positive axial spike on the x-axis and a negative half-axis, and the parabola functions as y=ax2, then the area formula is:

area = 1/3ax3 + b/2ax2 + cx。

where a, b, and c are the x-coordinates of the parabola at the intersection of the positive and semi-axes of the x-axis, the negative semi-axis of the x-axis, and the intersection of the parabola.