length = 40-2s
width = 35-2s
height = s
a) V = s(40-2s)(35-2s)
= ..
= 1400s - 150s^2 + 4s^3
b) standard shape of a cubic, with x-intercepts of
0 , 20, and 17.5
c) make special notice of where V is above and below the x-axis
d) dV/ds = 1400 - 300s + 12s^2
= 0 for a max of V
divide by 4 ...
3s^2 - 75s + 350 = 0
s = (75 ± √1425)/6
= appr 18.8 or appr 6.2
clearly the 18.8 would produce a negative width, so we reject that
accept s = 6.2 to yield a max volume of appr 3867 cubic inches
to get a volume of 1225 we set
4s^3 - 150s^2 + 1400s = 1225
4s^3 - 150s^2 + 1400s - 1225 = 0 gives a solution of
s = .974 , 13.9 and 22.6
by Wolfram:
http://www.wolframalpha.com/input/?i=4s%5E3+-+150s%5E2+%2B+1400s+-+1225%3D0
both .974 and 13.9 yield our needed result, while 22.6 would produce a negative length and width, thus we must reject it.
An open box is to be made from cutting squares of side 's' from each corner of a piece of cardboard that is 35" by 40".
(a) Write an expression for the volume, 'V', of the box in terms of 's'.
(b) Draw a graph of V(s).
(c) State the domain and range of V(s).
(d) Find the value of 's' that will give the maximum volume. What is the maximum volume?
(e) What realistic value(s) of 's' will generate a volume of 1225 cubic units?
1 answer
1 answer