Practice Problems for 4/15
- Let \(x\) and \(y\) represent two non-negative numbers whose sum is 9. Find \(x\) and \(y\) so that the product of one number and the square of the other is maximized.
Goal: Maximize the value of \(xy^2\)
Restrictions:
\( 0\leq x,y \leq 9\)
\(x+y=9\)
From the second restriction we have \(y=9-x\), so we want to find a maximum value of the function \(f(x)=x(9-x)^2\).
Find critical points: \(f'(x)=(9-x)^2-2x(9-x)=0\) if \(x=3\). Classify this critical point using the second derivative test: \(f''(x)=-2(9-x)-2(9-x)+2x\) and \(f''(3)<0\), which implies \(x=3\) is the location of a local maximum, and the local maximum value is \(f(3)=108\).
Since \(x\) is in a closed interval, we must also check the endpoints: \(f(0)=f(9)=0<108\), so \(108\) is the maximum value of the product.
- Find the function \(f(x)\) whose derivative is \(f'(x)=x^2-4\) and \(f(3)=-5\).
First, find the general antiderivative: \(f(x)=\frac{1}{3}x^3-4x+C\).
Second, use the given value of \(f\) to solve for \(C\): \(-5=f(3)=\frac{1}{3}3^3-4(3)+C \implies -5=-3+C\). It follows that \(C\) must be equal to -2 and so \(f(x)=\frac{1}{3}x^3-4x-2\).
- An impatient driver wants to pass the car in front of him. What constant acceleration, \(a\), is required to increase the speed of the passing car from 30 mph to 50 mph in 5 seconds?
First, convert the velocity into ft/second since we are interested in time units of seconds: 30 mph=44 ft/s, and 50 mph= 220/3 ft/s.
The acceleration at time \(t\) is going to be constant: \(a(t)=a\) ft/s.
Given:
v(0)=44
v(5)=220/3
v'(t)=a
From the last piece of information we know \(v(t)=at+C\), for some constant \(C\). Using this along with the first piece of given information we have \(44=v(0)=0+C\), so \(C=44\) and \(v(t)=at+44\). We are trying to find \(a\), so we must using the fact that \(v(5)=220/3\): \(220/3=v(5)=5a+44 \implies a=88/15\). So the constant acceleration must be \(\frac{88}{15}\) ft/s\(^2\).