The volume of a box(V) varies directly with its length(l). If a box in the group has a length of 30 inches, and the girth of 20 inches (perimeter of the side formed by the width and height), what is its height? Use k = 24. (Hint: Volume = length width height. Solve for length, and substitute into the equation for constant of proportionality.) ___ inches or ___ inches

Respuesta :

V = kl   where k is the  constant of proportionality

V = 24l = 24*30 = 720 in^3

720 = 30*h*w
hw = 720 / 30 = 24
also 2h + 2w = 20 
w + h = 10
24/h  + h = 10
24 + h^2 = 10h
h^2 - 10h + 24 = 0
(h - 4)(h - 6)=0 

h = 4 inches or 6 inches.

Answer:

Volume of box (V) is given by:

[tex]V = lwh[/tex]            .....[1]

where,

l is the length , w is the width and h is the height of the box respectively.

As per the statement:

The volume of a box(V) varies directly with its length(l).

[tex]V \propto l[/tex]

then;

[tex]V = kl[/tex] where, k is the constant of proportionality.

Substitute  k = 24 and l = 30 inches we have;

[tex]V = 24 \cdot 30 = 720 in^3[/tex]

Substitute the given values of V  and l in [1] we have;

[tex]720 = 30wh[/tex]

Divide both sides by 30h we have;

[tex]\frac{24}{h} =w[/tex]               .....[2]

It is also given that  the girth of 20 inches (perimeter of the side formed by the width and height)

Perimeter of rectangle formed by width and height is given by:

P = 2(w+h)

then;

[tex]20 = 2(w+h)[/tex]

Divide both sides by 2 we have;

[tex]10 = w+h[/tex]

or

w+h = 10                   .....[3]

Substitute equation [2] into [3] we have;

[tex]\frac{24}{h}+h = 10[/tex]

⇒[tex]24 +h^2 = 10h[/tex]

⇒[tex]h^2-10h+24 = 0[/tex]

⇒[tex]h^2-6h-4h+24=0[/tex]

⇒[tex]h(h-6)-4(h-6)=0[/tex]

Take h-6 common we have;

⇒[tex](h-6)(h-4)=0[/tex]

By zero product property we have;

h-6=0 or h-4=0

⇒h = 6 inches or h =  4 inches.

therefore, the height is,  6 or 4 inches