A wise old troll wants to make a small hut. Roofing material costs five dollars per square foot and wall materials cost three dollars per square foot. According to ancient troll customs the floor must be square, but the height is not restricted.
(a) Express the cost of the hut in terms of its height h and the length x of the side of the square floor. ($)
(b) If the troll has only 960 dollars to spend, what is the biggest volume hut he can build? (ft^3)

Respuesta :

Answer:

a) Cost (h,x)  =  12*x*h + 5*x²

b)

V =  V(max) = 355.5 ft³

Dimensions of the hut:

x = 9.48 ft       (side of the base square)

h = 3.95 ft      ( height of the hut)

Step-by-step explanation:

Let x be the side of the square of the base

h the height of the hut

Then the cost of the hut as a function of "x"  and "h" is

Cost of the hut = cost of 4 sides + cost of roof

cost of side  = 3* x*h   then for four sides cost is   12*x*h

cost of the roof = 5 * x²

Cost(h,x)  =  12*x*h + 5*x²

If the troll has only 900 $

900 = 12xh + 5x²        ⇒ 900 - 5x² = 12xh      ⇒(900-5x²)/12x  = h

And the volume of the hut   is  V  =  x²*h       then

V (h)  =  x² * [(900-5*x²)]/12x

V(h)  = x (900-5x²) /12      ⇒ V(h) = (900*x - 5*x³) /12

Taking derivatives (both sides of the equation):

V´(h) =  (900 - 10* x²)/12                V´(h) =  0

900 - 10*x² = 0          ⇒ x² = 90         x =√90      

x = 9.48 ft

And h

h = (900-5x²)/12x       ⇒  h = [900 - 90(5)]/12*x    ⇒ h = 450/113,76

h = 3.95 ft

And finally the volume of the hut is:

V(max) = x²*h             ⇒   V(max) = 90*3.95

V(max) = 355.5 ft³

A hut will consist of four walls and one roof. The figures needed evaluates to:

  • The cost of the hut expressed in terms of its height h and the length x of the side of the square floor is [tex]12hx + 5x^2 \text{\:(in dollars) }[/tex]
  • The biggest volume hut that can be build with 960 dollars at max is 426.67 sq. ft approx

How to find the volume of cuboid?

Let the three dimensions(height, length, width) be x, y,z units respectively.

Then the volume of the cuboid is given as

[tex]V = x \times y \times z \: \rm unit^3[/tex]

How to obtain the maximum value of a function?

To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.

Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.

For this case, we're specified that:

  • Cost of roofing material = $5 / sq. foot
  • Cost of wall material = $3 / sq. foot
  • The side length of floor = side length of roof = [tex]x \: \rm ft[/tex]
  • The height of the room = [tex]h \: \rm ft[/tex]

Four walls are attached to sides of floor. Thus, their one edge's length = length of side of floor = [tex]x \: \rm ft[/tex]

Thus, we get:

  • Area of four walls = [tex]4 \times (h \times x)[/tex] sq. ft

Thus, cost of four walls' material = [tex]3 \times 4 \times h \times x = \$ 12hx[/tex]

  • Area of roof = [tex]side^2 = x^2 \: \rm ft^2[/tex]

Thus, cost of roofing material for this roof = [tex]5 \times x^2 = \$5x^2[/tex]

Thus, cost of hut = cost for walls + cost for roof = [tex]12hx + 5x^2 = x(12h+5x) \: \rm (in \: dollars)[/tex]

The volume of the hut is: [tex]x\times x\times h =x^2.h \: \rm ft^3[/tex]

If troll has got only $960, then,

[tex]12hx + 5x^2 \leq 960\\\\\text{Multiplying x on both the sides}\\\\12x^2h + 5x^3 \leq 960x\\\\x^2h \leq \dfrac{960x-5x^3}{12}[/tex]

Let we take [tex]f(x) =\dfrac{960x-5x^3}{12}[/tex]

Then, taking its first and second derivative, we get:
[tex]f(x) =\dfrac{960x-5x^3}{12}\\\\f'(x) = 80 -1.25x^2\\\\f''(x) = -2.5x[/tex]

Putting first derivative = 0, we get critical points as:

[tex]80-1.25x^2 = 0\\\\x = 8 \text{\:(positive root as x denotes side length, thus a non-negative quantity)}[/tex]

At x = 8, the second derivative evaluates to:

[tex]f''(8) = -2.5(8) < 0[/tex]

Thus, we obtain maxima at x = 8.

Thus, we get the maximum value of function when x = 8.

Since we have:

[tex]V = x^2h \leq f(x)[/tex] (V is volume of the hut)

and [tex]max(f(x)) = f(8) = \dfrac{960(8) - 5(8)^3}{12} = 640-213.3\overline{3} \approx 426.67[/tex]

Thus, max(V) = 426.67 sq. ft approximately.

Thus, the figures needed evaluates to:

  • The cost of the hut expressed in terms of its height h and the length x of the side of the square floor is [tex]12hx + 5x^2 \text{\:(in dollars) }[/tex]
  • The biggest volume hut that can be build with 960 dollars at max is 426.67 sq. ft approx

Learn more about maxima and minima here:

https://brainly.com/question/13333267