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Chapter 7: Problem 61

If \(x+y=60, x>0, y>0\), then the expression \(x^{2}(30-y)^{2}\) has (A) least value \(=0\) (B) greatest value \(=15^{4}\) (C) two extrema (D) no greatest value

Short Answer

Expert verified
(A) The expression has the least value of 0. (B) The expression has the greatest value \(= 15^4\). (C) The expression has two extrema. (D) The expression does not have a greatest value. Answer: (B) and (C) are true.

Step by step solution

01

Rewrite the equation to isolate one variable

We are given the equation \(x+y=60\). In order to make the expression \(x^2 (30-y)^2\) easier to work with, we can rewrite the equation to isolate the variable \(y\). Solve for \(y\) in terms of \(x\): \(y = 60 - x\)
02

Substitute the new expression of y in the given expression

Now, let's substitute the expression of \(y\) we obtained in step 1 into \(x^2 (30-y)^2\) to obtain a new expression in terms of only \(x\): \(x^2 (30-(60 - x))^2 = x^2 (x-30)^2\)
03

Find the derivative of the new expression

In order to find the extrema points, we need to find the critical points where the derivative of our expression is equal to 0. So, we first find the derivative of \(x^2 (x-30)^2\) with respect to \(x\): \(\frac{d}{dx}(x^2 (x-30)^2) = 4x^3 - 180x^2 + 2700x\)
04

Find the critical points of the expression

Now, we want to find the critical points where the derivative is equal to 0: \(4x^3 - 180x^2 + 2700x = 0\) Simplify the equation by dividing by \(4x\): \(x^2 - 45x + 675 = 0\)
05

Solve quadratic equation for x

Solve the quadratic equation obtained in step 4 for the values of \(x\): \(x^2 - 45x + 675 = 0\) Using the quadratic formula, we get two real solutions: \(x=15\) and \(x=45\). However, based on the constraint \(x > 0\), we have two possible critical points: \(x=15\) and \(x=45\).
06

Determine the values of y and the extremum points

Using the two critical points found in step 5, we can determine the values of \(y\) and the extremum points: When \(x=15\), \(y= 60 - x = 45\). So, using the given expression, the first possible extremum is \(15^2 (15)^2 = 15^4\). When \(x=45\), \(y=60 - x = 15\). So, using the given expression, the second possible extremum is \(45^2 (15)^2\).
07

Analyze the Characteristics of the Expression

We found the expression's maximum value at \(x=45\) is \(45^2 (15)^2\). This means the expression does have a greatest value (\((D)\) is false). When \(x=15\) or \(y=45\), we found the extremum to be \(15^4\); thus, the expression has its least value at \(x=45\), but this least value is not 0 (\((A)\) is false). The greatest value indeed equals to \(15^4\) (\((B)\) is true). As for extrema, we found two critical points, \(x=15\) and \(x=45\), so the expression does have two extrema (\((C)\) is true). Therefore, the correct answer is: \(x^2 (30-y)^2\) has (B) greatest value \(=15^4\) and (C) two extrema.

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