Let’s solve
Problem 1 step by step, as requested.
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Problem 1:
> Find two positive numbers such that the sum of the first and twice the second is 100 and their product is a maximum.
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##
a. What is being maximized or minimized?
-
Words: We are trying to
maximize the product of two positive numbers.
-
Equation: Let the two numbers be \( x \) and \( y \).
→ The product is \( P = x \cdot y \)
✔ Answer for (a):
Words: Maximize the product of the two numbers.
Equation: \( P = xy \)
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b. What are the constraints? Do you need to draw a picture?
-
Words: The constraint is: “the sum of the first and twice the second is 100.”
Also, both numbers must be
positive.
-
Equation: \( x + 2y = 100 \)
-
Do you need to draw a picture? Not necessary for this algebraic optimization problem — it’s purely symbolic.
✔ Answer for (b):
Words: The sum of the first number and twice the second number equals 100; both numbers are positive.
Equation: \( x + 2y = 100 \), with \( x > 0 \), \( y > 0 \)
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c. Solve the constraint equation for one variable:
From \( x + 2y = 100 \), solve for \( x \):
\[
x = 100 - 2y
\]
✔ Answer for (c):
\( x = 100 - 2y \)
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d. Use the constraint equation to rewrite the max/min function in terms of one variable and simplify it.
We had \( P = xy \). Substitute \( x = 100 - 2y \):
\[
P(y) = (100 - 2y) \cdot y = 100y - 2y^2
\]
✔ Answer for (d):
\( P(y) = 100y - 2y^2 \)
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##
e. Find the critical points. Determine the absolute max or min.
Take derivative of \( P(y) \) with respect to \( y \):
\[
P'(y) = 100 - 4y
\]
Set derivative equal to zero:
\[
100 - 4y = 0 \Rightarrow y = 25
\]
Now find corresponding \( x \):
\[
x = 100 - 2(25) = 50
\]
Check if this is a maximum:
Second derivative: \( P''(y) = -4 < 0 \) → concave down →
maximum
Also, check domain: Since \( x > 0 \) and \( y > 0 \), from \( x = 100 - 2y > 0 \Rightarrow y < 50 \), and \( y > 0 \), so \( y \in (0, 50) \). Our critical point \( y = 25 \) is inside this interval.
✔ Answer for (e):
Critical point at \( y = 25 \), then \( x = 50 \).
Since \( P''(y) = -4 < 0 \), this is an
absolute maximum.
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f. Read the problem again, have you answered it? Does your answer make sense in the problem? Write a sentence to answer the question.
Yes, we found two positive numbers:
50 and 25, such that:
- \( x + 2y = 50 + 2(25) = 100 \)
✔
- Their product is \( 50 \cdot 25 = 1250 \), which is the maximum possible under the constraint.
This makes sense because if you try other values (like \( y = 20 \), then \( x = 60 \), product = 1200 < 1250; or \( y = 30 \), \( x = 40 \), product = 1200 < 1250), the product is smaller.
✔ Answer for (f):
The two positive numbers are
50 and 25, and their product of
1250 is the maximum possible under the given constraint.
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##
✔ Final Answer Summary for Problem 1:
The two numbers are 50 and 25.
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If you’d like me to solve Problems 2, 3, and 4 next, just say the word! But since you only uploaded
one image with these problems, I’ve solved the first one completely as requested.
Let me know if you want help with the others too!
Parent Tip: Review the logic above to help your child master the concept of calculus optimization worksheet.
