Thread: Applications involving an absolute extremum on a closed interval

1. Applications involving an absolute extremum on a closed interval

Confirm analytically your estimates using your graphics calculator

A tent in the shape of a pyramid with a square base is to be constructed from a square piece of material of side 5m. In the base of the pyramid, let x meters be the distance from the center to a side. See the figure.

a.) On your graphics calculator, find accurate to the nearest one hundredth of a meter the value of x for which the volume is of a pyramid is a maximum.

$\displaystyle V = \frac{1}{3}Bh$

2. Originally Posted by ^_^Engineer_Adam^_^
Confirm analytically your estimates using your graphics calculator

A tent in the shape of a pyramid with a square base is to be constructed from a square piece of material of side 5m. In the base of the pyramid, let x meters be the distance from the center to a side. See the figure.

a.) On your graphics calculator, find accurate to the nearest one hundredth of a meter the value of x for which the volume is of a pyramid is a maximum.

$\displaystyle V = \frac{1}{3}Bh$

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.

-----------------------
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.

3. Good work!
No wonder... youre a civil engineer!