Wize High School Grade 12 Calculus Textbook > Derivative Applications

Maximum & Minimum on a Closed Interval

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Absolute Max & Min
Given a function f(x)f\left(x\right)f(x) on a closed interval a≤x≤ba\le x\le ba≤x≤b, 
the global (absolute) maximum value is the largest value on this interval
the global (absolute) minimum value is the smallest value on this interval
The maximum and minimum values of a function on a closed interval are called extreme values or absolute extrema.

How do we find the maximum and minimum values?
A max or min value of f(x)f(x)f(x) on a closed interval [a, b]\left[a,\ b\right][a, b]must occur at
at a turning point when f′(x)=0f'\left(x\right)=0f′(x)=0
an endpoint (i.e. x=ax=ax=a or x=bx=bx=b)
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Example
Determine the absolute extrema of f(x)=x33−x2f\left(x\right)=\frac{x^3}{3}-x^2f(x)=3x3​−x2 on the interval −2≤x≤3-2\le x\le3−2≤x≤3.

Step 1: Find the derivative and set it equal 0
f′(x)=3x23−2xf'\left(x\right)=\frac{3x^2}{3}-2xf′(x)=33x2​−2x
f′(x)=x2−2xf'\left(x\right)=x^2-2xf′(x)=x2−2x
0=x2−2x0=x^2-2x0=x2−2x
0=x(x−2)0=x\left(x-2\right)0=x(x−2)
x=0 or 2x=0\ or\ 2x=0 or 2

Step 2: Evaluate the function values at the points found in step 1 and at the endpoints
f(0)=033−02=0f\left(0\right)=\frac{0^3}{3}-0^2=0f(0)=303​−02=0
f(2)=233−22=83−4=−43f\left(2\right)=\frac{2^3}{3}-2^2=\frac{8}{3}-4=-\frac{4}{3}f(2)=323​−22=38​−4=−34​
f(−2)=(−2)33−(−2)2=−83−4=−203f\left(-2\right)=\frac{\left(-2\right)^3}{3}-\left(-2\right)^2=-\frac{8}{3}-4=-\frac{20}{3}f(−2)=3(−2)3​−(−2)2=−38​−4=−320​
f(3)=333−32=9−9=0f\left(3\right)=\frac{3^3}{3}-3^2=9-9=0f(3)=333​−32=9−9=0

Step 3: Compare these values and make a conclusion
The smallest of these values found in step 2 is f(−2)=−203f\left(-2\right)=-\frac{20}{3}f(−2)=−320​.
The largest of these values found in step 2 are f(0)=0f\left(0\right)=0f(0)=0 and f(3)=0f\left(3\right)=0f(3)=0

Therefore, the absolute minimum on this interval is −203-\frac{20}{3}−320​ at x=−2x=-2x=−2, the absolute maximum on this interval is 000 at the points x=0, 3x=0,\ 3x=0, 3

Practice: Max & Min
Find the absolute extrema of the following functions on the given interval.

f(x)=(x−3)2f\left(x\right)=\left(x-3\right)^2f(x)=(x−3)2 on the interval 0≤x≤30\le x\le30≤x≤3

The absolute maximum

f

value =

The absolute maximum value occurs at

x=

The absolute minimum

f

value =

The absolute minimum value occurs at

x=

I don't know

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Example: Max & Min Application Problem
The cost of production (in $) at a company is modelled by C(x)=1000+10x+0.05x2C\left(x\right)=1000+10x+0.05x^2C(x)=1000+10x+0.05x2, where the units produced is 1≤x≤2001\le x\le2001≤x≤200. How many units must this company produce to minimize the unit cost U(x)=C(x)xU\left(x\right)=\frac{C\left(x\right)}{x}U(x)=xC(x)​?

The unit cost is U(x)=1000+10x+0.05x2x=1000x+10+0.05xU\left(x\right)=\frac{1000+10x+0.05x^2}{x}=\frac{1000}{x}+10+0.05xU(x)=x1000+10x+0.05x2​=x1000​+10+0.05x.
We want to find the minimum value of this function:

Step 1. Find the derivative and set to 0
U(x)=1000x−1+10+0.05xU\left(x\right)=1000x^{-1}+10+0.05xU(x)=1000x−1+10+0.05x
U′(x)=−1000x−2+0.05U'\left(x\right)=-1000x^{-2}+0.05U′(x)=−1000x−2+0.05
0=−1000x2+0.050=-\frac{1000}{x^2}+0.050=−x21000​+0.05
−0.05=−1000x2-0.05=-\frac{1000}{x^2}−0.05=−x21000​
x2=20000x^2=20000x2=20000
x≈141.42x\approx141.42x≈141.42

Step 2. Evaluate the function values
U(141.42)=1000141.42+10+0.05(141.42)=24.14U\left(141.42\right)=\frac{1000}{141.42}+10+0.05\left(141.42\right)=24.14U(141.42)=141.421000​+10+0.05(141.42)=24.14
U(1)=10001+10+0.05(1)=1010.05U\left(1\right)=\frac{1000}{1}+10+0.05\left(1\right)=1010.05U(1)=11000​+10+0.05(1)=1010.05
U(200)=1000200+10+0.05(200)=25U\left(200\right)=\frac{1000}{200}+10+0.05\left(200\right)=25U(200)=2001000​+10+0.05(200)=25

Step 3. Compare values
Therefore, the unit cost is minimized at approximately 141.42 units. However, since we cannot produce partial units, we either have to product 141 or 142 products:
U(141)=1000141+10+0.05(141)≈24.14220U\left(141\right)=\frac{1000}{141}+10+0.05\left(141\right)\approx24.14220U(141)=1411000​+10+0.05(141)≈24.14220
U(142)=1000142+10+0.05(142)≈24.14225U\left(142\right)=\frac{1000}{142}+10+0.05\left(142\right)\approx24.14225U(142)=1421000​+10+0.05(142)≈24.14225
Therefore, 141 units will minimize unit cost.

Practice: Max & Min Application Problem
The mass of a certain sample of bacteria is modelled by A(t)=t+1tA\left(t\right)=\sqrt{t}+\frac{1}{t}A(t)=t​+t1​, 1≤t≤251\le t\le251≤t≤25, where ttt is in hours.

Find the absolute maximum and minimum mass of bacteria on this interval.

Enter the absolute maximum mass of bacteria, rounded to 2 decimal places

Enter the absolute minimum mass of bacteria, rounded to 2 decimal places

I don't know

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Example: Max & Min of a Function
The function f(x)=ax2+bx+3f\left(x\right)=ax^2+bx+3f(x)=ax2+bx+3 on the interval x∈[0, 7]x\in\left[0,\ 7\right]x∈[0, 7] has an absolute maximum value of 242424 at x=7x=7x=7, and an absolute minimum value at x=2x=2x=2. Find the values of aaa and bbb.

The absolute max and min must occur when f′(x)=0f'\left(x\right)=0f′(x)=0 or at the endpoints x=0 or 7x=0\ or\ 7x=0 or 7. Since the absolute minimum value is at x=2x=2x=2, this must be the point where the derivative f′(x)=0f'\left(x\right)=0f′(x)=0

Find the derivative:
f′(x)=2ax+bf'\left(x\right)=2ax+bf′(x)=2ax+b

Sub in x=0x=0x=0 and f′=0f'=0f′=0
0=4a+b\boxed{0=4a+b}0=4a+b​ - equation 1

We also know that the maximum value is 24 at x=7x=7x=7:
24=a(72)+b(7)+324=a\left(7^2\right)+b\left(7\right)+324=a(72)+b(7)+3
21=49a+7b21=49a+7b21=49a+7b
3=7a+b\boxed{3=7a+b}3=7a+b​ - eqution 2

Now let's solve for a and b:
Equation 1: b=−4ab=-4ab=−4a
Sub this into equation 2: 3=7a+(−4a) → 3=3a →a=13=7a+(-4a)\ \to\ 3=3a\ \to a=13=7a+(−4a) → 3=3a →a=1
Solve for bbb: b=−4(1)=−4b=-4\left(1\right)=-4b=−4(1)=−4

Therefore, a=1a=1a=1 and b=−4b=-4b=−4.