Expert: Abe Mantell - 1/23/2005

Question
I have three Calculus Questions that I have worked on for three hours and I can't figure them out.  I don't understand the process of finding the answers.  Can you help me?

Question 1:
A tank with a rectangular base and rectangular sides is to be open at the top.  It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters.  If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?

Question 2:
Find the maximum volume of a box that can be made by cutting out squares from the corners of an 8-inch by 15-inch rectangular sheet of cardboard and folding up the sides.  Justify your answer.

Question 3:
Find the area of the largest rectangle (with the sides parallel to the coordinate axes) that can be inscribed in the region enclosed by the graphs of f(x)=18-x^2 and g(x)=2x^2-9.

If you can help me out that would be wonderful, thank you for your time.

Answer
Hello Ashley,

1. Let x=length & y=depth, with 4=width, and volume, V=36
-- so, 4xy=36 ==> y=9/x
-- the area of the sides is 2(xy)+2(4y)=2xy+8y=2x(9/x)+8(9/x)
-- = 18+72/x
-- the area of the base is 4x
-- thus, the total cost is: C(x)=$10*4x+$5*(18+72/x)
-- C(x)=40x+90+360/x
-- now solve C'(x)=0

2. let x=length of the square "cut-out"
-- so the volume of the box will be, V=(8-2x)(15-2x)
-- now solve  V'=0

3. The two parabolas intersect at x=-3 & x=+3
-- let x=the x-value of the width of the rectangle,
-- so the entire width is 2x
-- the height is (18-x^2)-(2x^2-9)=27-3x^2
-- so the area is A(x)=2x(27-3x^2)
-- now solve for A'(x)=0

OK?  Can you finish them off from here?
Let me know if you need help.

TTYL, Abe