Answered

Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

PRACTICE ANOTHER A piece of wire 18 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (a) How much wire should be used for the square in order to maximize the total area? m (b) How much wire should be used for the square in order to minimize the total area? m

Sagot :

Answer:

Step-by-step explanation:

We have the equations

4x + 3y = 18   where x = the side of the square and y = the side of the triangle

For the areas:

A = x^2 + √3y/2* y/2

A = x^2  + √3y^2/4

From the first equation x = (18 - 3y)/4

So substituting in the area equation:

A = [ (18 - 3y)/4]^2 + √3y^2/4

A = (18 - 3y)^2 / 16 + √3y^2/4

Now for maximum / minimum area the derivative = 0 so we have

A' = 1/16 * 2(18 - 3y) * -3 + 1/4 * 2√3 y = 0

-3/8 (18 - 3y) + √3 y /2 = 0

-27/4 + 9y/8 + √3y /2 = 0

-54 + 9y + 4√3y = 0

y = 54 / 15.93

= 3.39 metres

So x = (18-3(3.39) / 4 = 1.96.

This is a minimum value for x.

So the total length of wire the square  for minimum  total area is 4 * 1.96

= 7.84 m

There is no maximum area as the equation for the total area is a quadratic with a positive leading coefficient.

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.