Differentiation : Critical Point - Find Maximum Value

A manufacturer produces cardboard boxes that are open at the top and sealed at the base. The base is rectangular and its length is double its width. Let x denote the width in metres. the surface area of each such box is fixed to be 3 square metres. The manufacturer wishes to determine the height h and the base width x, in metres, of the box so that its volume is as large as possible.

a) Express the volume V in terms of x and h
b) Show that 2x^2+6xh =3 (consider the surface area)
c) Express h in terms of x and deduce that
V=x-(2/3)x^3
d) Use calculus to find the value of x so that V is as large as possible. Justify your answer. What is the largest possible value of the volume?

Solution Summary

The maximum volume is found using a critical point. The solution is detailed.

... 9. Using the First and Second Derivative tests as appropriate ... To find critical points, put h'(x) = 0. ... at x = -2 and its value is -1 and local maximum value at x ...

...Find the first order condition for a critical point of this ... a maximum or a minimum or an inflection point? ... minimum we check the sign of the second derivative: ...

... Note: it is also possible to combine this technique with the first and second derivative tests to find which critical points are local minima and ...

... 3. Use information from the derivative of each function to help you graph the function. ... To find critical points, put g'(x) = 0. Critical points are x = 0, 5. ...

... to find all critical points and use the second derivative to find all inflections ... Use a graph to identify each critical point as a local maximum, a local ...

... Let us first find the critical points for the given function for that find the first ... Therefore by second derivative test Q = 50 will be maximum point of the ...

... 3. Find dy/dx by implicit differentiation. ... 4. Determine the critical numbers for the given function and classify each critical point as a relative maximum...