2. (#8 on exam 3 review) the sum of two numbers is 8. find the numbers if the sum of their cubes is to be…

Explanation:

Step1: Define variables

Let one number be (x), then the other number is (y = 8 - x). Let (S) be the sum of their cubes. So (S=x^{3}+y^{3}=x^{3}+(8 - x)^{3}).

Step2: Expand the function

Expand ((8 - x)^{3}) using the formula ((a - b)^{3}=a^{3}-3a^{2}b + 3ab^{2}-b^{3}), where (a = 8) and (b=x). ((8 - x)^{3}=8^{3}-3\times8^{2}x+3\times8x^{2}-x^{3}=512-192x + 24x^{2}-x^{3}). Then (S=x^{3}+512-192x + 24x^{2}-x^{3}=24x^{2}-192x + 512).

Step3: Find the first - derivative

Differentiate (S) with respect to (x). Using the power rule (\frac{d}{dx}(ax^{n})=nax^{n - 1}), where (a = 24), (n = 2) for the first term, (a=-192), (n = 1) for the second term and the derivative of a constant (512) is (0). (S^\prime=\frac{dS}{dx}=48x-192).

Step4: Find the critical points

Set (S^\prime = 0). (48x-192 = 0). Add (192) to both sides: (48x=192). Divide both sides by (48): (x = 4).

Step5: Find the second - derivative

Differentiate (S^\prime) with respect to (x). (\frac{d}{dx}(48x-192)=48). Since (S^{\prime\prime}=48>0), the function (S) has a minimum at (x = 4). When (x = 4), (y=8 - 4=4).

Answer:

The two numbers are (4) and (4).