Calculus I Max/Min Problem Practice

This practice exercise covers
Solving Optimization Word Problems

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Make a diagram for each question.

QUESTIONS

1) For the given cost function C(x) and demand function p(x),
find the production level x that will maximize the profit P(x).

a) C(x) = 680 + 4x + 0.01x², and p(x) = 12.
b) C(x) = 1450 + 36x - x² + 0.001x³, and p(x) = 60 - 0.01x.

2) If 3 sides of a trapezoid measure 10 cm., how long must the
4th side be to maximize the area of the figure?

3) From a piece of cardboard with area 400 sq. cm. we want to make a box
with a square base and no top. Find the dimensions that maximize the volume.

4) A printed page with 2 inch margins top and bottom; 1 inch margins on the sides
is to have a printed area of 18 in². Find the dimensions of the page that minimize its area.

(solution)

5) An 8 cm. square piece of cardboard will be used to create an open box by
cutting equal squares from the corners and folding up the sides. What is the maximum
volume of such a box?

6) Find the area of the largest rectangle that can be inscribed in a right triangle
with base 10 cm. and height 8 cm..

7) Find the minimum value for the perimeter of a rectangle with area = 100 cm².

8) Find 2 positive real numbers whose sum is 20 and whose product is a maximum.

9) Separate 40 into two parts such that the sum of the squares of the parts is minimized.

The dreaded "wire" question. Every Calculus book has one:

10) A piece of wire 24 cm. long is cut into 2 pieces, one of which is formed
into a circle, the other is used to form a square. Where should we cut the wire
to minimize the area enclosed by both the circle and square?

SOLUTIONS

 1) a) P(x) = R(x) - C(x) R (x) = 12x and C(x) = 4x + 0.01x² + 680 so P(x) = 8x - 0.01x² - 680 which makes P '(x) = 8x - 0.02x = 0 max profit at production level of x = 400.b) R(x) = 60x - 0.01x² C(x) = 36x - x²+ 0.001x³ + 1450 P(x) = -1450 + 24x + 0.99x² - 0.001x³ so P '(x) = 24 + 1.98x - 0.003x² = 0 using the quadratic formula, x = 671.91. max profit at production level of x = 672. 2) From the diagram, Area of a trapezoid is ½ h(b1 + b2) So,  multiply through by and solve the quadratic 50 - 5x - x² = 0. maximum area at x = 5. The 4th side = 20 cm. 3 The surface area must be 400 cm². So SA = x² + 4xh = 400. Solve for h.    4) The page area is Simplified, So If x = 3, the page is 5 in. by 10 in. 5) 6) with similar triangles, find y in terms of x. Now So make the base 5 cm and the height 4 cm. The maximum area is 20 cm². 7)   The minimum perimeter is 4(10) = 40 cm. 8)x + y = 20 which gives y = 20 - x P = x(20 - x) = 20x - x² P ' = 20 - 2x u x = 10.9) minimize x² + (40 - x)² = f (x) expand to get f (x) = 2x² - 80x + 200 a parabola, opens up with a min so f '(x) = 4x -80 = 0 minimum at x = 20 and 40 - x = 20. 10) Since x is the circumference of the circle, x = 2o r This makes r = x / 2o Area of the circle = o r² = x² / 4oThe square's perimeter = 24 - x so each side is 6 - ¼ x. Area of the square = (6 - ¼ x)². Total Area function: A = x² / 4o + (6 - ¼ x)².Now, .

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