Sorry, but I do not know how to use any graphing calculator ---no kidding---so can I solve this analytically instead?

V = (1/3)Bh

means Volume = (1/3)(area of square base)(vertical height of the apex)

We know B = (2x)^2 = 4x^2

But we don't know h.

If we put up the pyramidal tent, the h is the vertical component of the slant height of the apex.

Let y = that slant height.

On the posted drawing, or when the tent is still flat on the ground or not yet cut, if we draw a horizontal, or vertical, line through the center of the 5m-square material, we see that

y +2x +y = 5

So,

2y = 5 -2x

y = (2.5 -x) m. ----------***

Put up the tent again.

We see an isosceles triangle silhouette. The base is 2x m long. The two equal sides are y meters each. Draw the h.

By Pythagorean Theorem,

y^2 = x^2 +h^2

h = sqrt(y^2 -x^2)

So,

h = sqrt[(2.5 -x)^2 -x^2]

h = sqrt[6.25 -5x +x^2 -x^2]

h = sqrt(6.25 -5x) ---------------***

Hence,

V = (1/3)(4x^2)sqrt(6.25 -5x)

V = (4/3)(x^2)sqrt(6.25 -5x)

Differentiate both sides with respect to x,

dV/dx = (4/3){(x^2)[(1/2)(-5 /sqrt(6.25 -5x)] +sqrt(6.25 -5x)[2x]}

dV/dx = (-10/3)(x^2)/sqrt(6.25 -5x) +(8/3)(x)sqrt(6.25 -5x)

Set dV/dx to zero,

0 = (-10/3)(x^2)/sqrt(6.25 -5x) +(8/3)(x)sqrt(6.25 -5x)

Multiply both sides by 3sqrt(6.25 -5x),

0 = -10x^2 +8(6.25 -5x)

0 = -5x^2 +4(6.25 -5x)

0 = -5x^2 +25 -20x

0 = -x^2 -4x +5

x^2 +4x -5 = 0

(x+5)(x-1) = 0

x = -5m or 1m

There are no negative dimensionsn so reject x = -5m

So, x = 1 m.

Therefore for maximum volume of the tent, x = 1 meter. -------answer.

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Ooppss, mistake. Let me edit it

....

0 = (-10/3)(x^2)/sqrt(6.25 -5x) +(8/3)(x)sqrt(6.25 -5x)

Multiply both sides by 3sqrt(6.25 -5x),

0 = -10x^2 +8x(6.25 -5x)

Divide both sides by 2x,

0 = -5x +4(6.25 -5x)

0 = -5x +25 -20x

0 = -25x +25 -------------------------:-), yeah, really?

x = 1 m ---------------hey!

Therefore for maximum volume of the tent, x = 1 meter. -------answer.