Problema Solution
the height and weight of a rectangular prism are each 2 inch shorter then the length of the prism. The volume of the prism is 40 cubic inches. Approximate the dimensions of the prism to the nearest hundredth
Answer provided by our tutors
let
l = the length of the prism
the height and width of a rectangular prism are each 2 inch shorter then the length of the prism
h = l - 2 the height of the prism
w = l - 2 the width of the prism
the volume of the prism is calculated by the formula V = l*h*w
l(l - 2)(l - 2) = 40
if we check the graph of the function y = x(x - 2)(x - 2) - 40 we see that the x-intercept is between 4 and 5
click here to see the graph
to fine the approximation we will use the Bisection method:
When f(x) is continuous in the environment of a single root r, then f(x) changes sign through r. There is a small interval [a,b] including r such that f(a).f(b) < 0. Taking the midpoint m of [a,b], there are three possibilities.
f(m) = 0 ; then m is the root r.
f(m).f(a) < 0 ; then the root r is in [a,m]
f(m).f(b) < 0 ; then the root r is in [m,b]
Now we can restart the procedure with the smaller interval [a,m] or [m,b]. The interval becomes smaller and smaller, so we can find an approximation of the root r.
In our case f(x) = x^3 - 4x^2 + 4x - 40
f(4) < 0, f(5) > 0 thus f(4)*f(5) < 0 and the start interval is
[4, 5]
(4 + 5)/2 = 4.5
f(4.5) < 0 thus the new interval is
[4.5, 5]
(4.5 + 5)/2 = 4.75
f(4.75) < 0
the new interval is
[4.75, 5]
and so on we find x = 4.86 to the nearest hundredth.
the the dimensions of the prism to the nearest hundredth are
l = 4.86 in
h = w = 4.86 - 2 = 4.66 in
let
l = the length of the prism
the height and width of a rectangular prism are each 2 inch shorter then the length of the prism
h = l - 2 the height of the prism
w = l - 2 the width of the prism
the volume of the prism is calculated by the formula V = l*h*w
l(l - 2)(l - 2) = 40
if we check the graph of the function y = x(x - 2)(x - 2) - 40 we see that the x-intercept is between 4 and 5
click here to see the graph
to fine the approximation we will use the Bisection method:
When f(x) is continuous in the environment of a single root r, then f(x) changes sign through r. There is a small interval [a,b] including r such that f(a).f(b) < 0. Taking the midpoint m of [a,b], there are three possibilities.
f(m) = 0 ; then m is the root r.
f(m).f(a) < 0 ; then the root r is in [a,m]
f(m).f(b) < 0 ; then the root r is in [m,b]
Now we can restart the procedure with the smaller interval [a,m] or [m,b]. The interval becomes smaller and smaller, so we can find an approximation of the root r.
In our case f(x) = x^3 - 4x^2 + 4x - 40
f(4) < 0, f(5) > 0 thus f(4)*f(5) < 0 and the start interval is
[4, 5]
(4 + 5)/2 = 4.5
f(4.5) < 0 thus the new interval is
[4.5, 5]
(4.5 + 5)/2 = 4.75
f(4.75) < 0
the new interval is
[4.75, 5]
and so on we find x = 4.86 to the nearest hundredth.
the the dimensions of the prism to the nearest hundredth are
l = 4.86 in
h = w = 4.86 - 2 = 2.86 in