10.1 (e) Solve optimization problems Lecture 3 0f 3
Strategy for solving optimum problems 1. Draw a diagram. Label the parts that are important in the problem. 2. Write an equation for the quantity whose maximum or minimum you want. 2
3. Express the quantity as a function in a single variable. 4. Differentiate to get f ’ (x), then f ’ (x)=0 to find all the x coordinates of stationary points. 5. Differentiate to get f” (x) to test the stationary points, whether it is maximum or minimum. 3
Example 1 Find two positive numbers whose sum is 30 and whose product is maximum. Solution let numbers are x and y x + y = 30 y = 30 - x so, their product is 2 (30 ) 30 P xy P x x x x
dx dp = 30 – 2x 30 – 2x = 0 2x = 30 x = 15 p is maximum, the numbers are x = 15 and y = 30 – 15 = 15 2 2 15, 2 0 d p when x dx
A cuboid has the following dimensions, length = ( 12 – 5x ) cm, breadth = x cm and height = 2x cm. Calculate the maximum volume of the cuboid. Example 2 Solution x cm 12 - 5x cm 2x cm
2 48x 30x dx dV 2 3 V x x 24 10 V x x x ( )(2 )(12 5 ) for a maximum volume, 0 dx dV 2 2 48 30 0 8 5 0 x x x x x x (8 5 ) 0 8 0 rejected 5 x or x x dx d V 48 60 2 2 when , , is maximumd V x dx d V V dx 2 2 2 2 8 848 605 5480 2 3 3max 8 8 24 10 20.485 5 V cm
A cone has a circular base with radius r and the height of the cone is h. Given that the slant height is , find the value of r and h if the volume of the cone is to be maximized. Example 3 48 h r O Solution V r h 2 3 1 volumeof cone, 3323116 483 1 483 1 h hhhhh 48 cm 2 + ℎ 2 = 48 2 2 = 48 − ℎ 2
0 dh dV for maximum volume, 2 2 2 16 0 , 16 0 16 4 or 4 h h h h rejected ℎ = 16 − ℎ 2 h dh d V 2 2 2 2 ℎ 2 = −2 4 = −8 < 0 ∴ V is maximum when h = 4cm
when h 4 48 16 2 r 4 2 32 2 r r r 4 2 since r is a radius the volume of the cone is maximum when r 4 2 cm and h 4 cm
10.2 RATE OF CHANGE OBJECTIVE a) Solve problems regarding rate of change including related rates.
*If y = f(x) ,then is the rate of change of y with respect to x dy dx * If the rate of change is negative, it means the quantity is decreasing with time. *If the rate of change is positive, it means the quantity is increasing with time. *the rate of change of quantity Q , means , the rate of change of Q with respect to time t. dQ dt
A cube has side of x cm. Estimate the rate in the total surface area of the cube as the side increases at a rate of 5 cm/s, when the side is 12 cm. EXAMPLE 4 Solution 5 / , dx Given cm s dt Total surface area of a cube , 2 6 12 12 12 144 A x dA x dx 2 144 5 720 /dAdt cm s = × 12 , ? , dA dA dA dxx cmdt dt dx dt
EXAMPLE 5 A cylindrical tank has a base with radius , 3 m.The tank is filled with water at the rate of 4 m3 /min. Find the rate of change of the water level in the tank. Given 3 , ? , dh dh dh dVr mdt dt dV dt 3 4 / min, dVmdt 9 dV dh Let V be the volume of the water in the tank. 2 V r h h r 2 (3 )h 9 h Solution dh dh dV dt dV dt 1 1 9 dh dV dVdh 149 4 / min9m
12cm 6cm The above figure shows a right circular cone with radius 6cm and height 12cm. An oil is pumped into the tank with the fixed rate of 10 cm3 /sec. Find the rate of change of height of the oil when the oil level is 5 cm deep. EXAMPLE 6
Let V be the volume of oil in the tank. 2 2 3 1 3 1 3 4 12 V r h h h h 6 12 r h 2 h r 2 2 4 h r Solution Given 3 10 / , dV cm s dt 5 , ? , dh dh dh dVh cmdt dt dV dt 2 4 dV h dh 410258 / 5dhdt cm s ℎ = ℎ × 25425 4dVdh