Practice problems for these solutions can be found at: Applications of Derivatives Practice Problems for AP Calculus

**Part A The use of a calculator is not allowed.**

- Volume: V =
- Pythagorean Theorem yields
- See Figure 8.6-1. Using similar triangles, with
*y*the length of the shadow you have: - See Figure 8.6-2. Volume of a cone
- See Figure 8.6-3.
- See Figure 8.6-4.
- See Figure 8.6-5.
- See Figure 8.6-5.
- See Figure 8.6-6.
- Step 1. Average Cost:
- See Figure 8.6-7.
- Step 1. Let
*h*be the height of the trough and 4 be a side of one of the two equilateral triangles. Thus, in a 30–60 right triangle, - See Figure 8.6-8.
- See Figure 8.6-9.
- Step 1. Let x be the distance of the foot of the ladder from the higher wall. Let
*y*be the height of the point where the ladder touches the higher wall. The slope of the ladder is - Step 1.
- See Figure 8.6-11.
- See Figure 8.6-12.

Surface Area: S=

*x*^{2} + *y*^{2} =(13)^{2}

At *x* =5, (5)^{2} + *y*^{2} =13^{2} y = ±12,

since *y* > 0, *y* =12.

Therefore,

Differentiate:

Using similar triangles, you have

reducing the equation to

Step 1. Using the Pythagorean Theorem, you have *x*^{2} + *y*^{2} =*z*^{2}. You also have

Step 2. Differentiate:

At *x* =120, both cars have traveled 3 hours and thus, *y* =3(30)=90. By the Pythagorean Theorem, (120)^{2} +(90)^{2} =z*2* z =150.

Step 3. Substitute all known values into the equation:

Step 4. The distance between the two cars is increasing at 50 mph at *x* =120.

Step 1. Applying the Pythagorean Theorem, you have

Step 2.

Step 3.

Step 4. First Derivative Test:

Thus at *x* =6, the area *A* is a relative maximum.

Step 5. Check endpoints. The domain of *A* is [9/2, 9]*A*(9/2)=0; and *A*(9)=0. Therefore, the maximum area of an isosceles triangle with the perimeter of 18 cm is cm^{2}. (Note that at *x* =6, the triangle is an equilateral triangle.)

Step 1. Let *x* be the number and be its reciprocal.

Step 2. *s* = with 0 < x < 2.

Step 3.

Step 4.

Step 5. Thus at *x* =1, *s* is a relative minimum. Since it is the only relative extremum, thus, at *x* =1, it is the absolute minimum.

Step 1. Volume: *V* =*x* (8–2*x* )(15–2*x*) with 0 ≤ *x* ≤ 4.

Step 2. Differentiate: Rewrite as *V* =4*x*^{3} – 46*x*^{2} +120*x*

Step 3. Set *V* =0 12*x*^{2} – 92*x* + 120=0 3*x*^{2} – 23*x* +30=0. Using the quadratic formula, you have defined for all real numbers.

Step 4. Second Derivative Test:

Step 5. Check endpoints. At *x* =0, *V* =0 and at *x* =4, *V* =0. Therefore, at *x* = *V* is the absolute maximum.

Step 1. Distance Formula:

Step 2. Let *S* = *Z*^{2}, since *S* and *Z* have the same maximums and minimums.

Step 3.

Step 4. Second Derivative Test:

Thus at *x* =1, *Z* has a minimum, and since it is the only relative extremum, it is the absolute minimum.

Step 5. At *x* =1,

Step 2.

Step 3. at *x* =0 which is not in the domain.

Step 4. Second Derivative Test:

Thus at *x* =2, the average cost is a minimum.

Step 1. Area:

*A*=

*x*(200 – 2

*x*)=200

*x*– 2

*x*

^{2}with 0 ≤

*x*≤ 100

Step 2. *A* '(*x* )=200 – 4*x*

Step 3. Set *A* '(*x* )=0 200 – 4*x* =0; *x* =50.

Step 4. Second Derivative Test:

*A*'(

*x*)= – 4; thus at

*x*=50, the area is a relative maximum.

*A*(50)=5000 m

^{2}.

Step 5. Check endpoints.

*A*(0)=0 and

*A*(100)=0; therefore at

*x*=50, the area is the absolute maximum and 5000 m

^{2}is the maximum area.

### Part B Calculators are allowed.

Step 2. Volume:

Step 3. Differentiate with respect to *t*.

Step 4. Substitute known values:

Step 1. tan θ = *S*/3000

Step 2. Differentiate with respect to *t*.

Step 3. At *t* =5; *S* =100(5)^{2} =2500; Thus, *Z*^{2} =(3000)^{2} +(2500)^{2} = 15,250,000. Therefore, *Z* = ±500 , since *Z* > 0, *Z* =500 . Substitute known values into the question:

=0.197 radian/sec. The angle of elevation is changing at 0.197 radian/sec, 5 seconds after lift off.

The water level in the cylinder is rising at

Step 2. Phythagorean Theorem:

Step 3. Enter The graph of *y*_{1} is continuous on the interval *x* > 8. Use the [*Minimum*] function of the calculator and obtain *x* =14.604; *y* =19.731. Thus the minimum value of *l* is 19.731 or the shortest ladder is approximately 19.731 feet.

Step 2. Enter:

Step 3. Use the [*Minimum*] function in the calculator and obtain *x* =790.6.

Step 4. Verify the result with First Derivative Test. Enter *y*2= *d*(2500/*x* +.02+004*x*, *x*); Use the [*Zero*] function and

Apply the First Derivative Test:

Thus the minimum average cost per unit occurs atx=790.6. (The graph of the average cost function is shown in Figure 8.6-10.)

Step 1. Area *A* =(2*x* )(2*y* ); 0 ≤ *x* ≤ 5 and 0 ≤ *y* ≤ 2.

Step 2. 4*x*^{2} +25*y*^{2} =100; 25*y*^{2} =100 – 4*x*^{2}.

Since *y*≥ 0

Step 3.

Step 4.

Step 5. Verify the result with the First Derivative Test. Enter

Use the [*Zero*] function and obtain *x* =3.536. Note that:

The function *f* has only one relative extremum. Thus it is the absolute extremum. Therefore, at *x* =3.536, the area is 20 and the area is the absolute maxima.

Step 1. Distance formula:

*l*

^{2}=

*x*

^{2}+

*y*

^{2};

*x*> 0.5 and

*y*> 4

Step 2. The slope of the hypotenuse:

Step 3.

Step 4. Enter y 1= and use the [*Minimum*] function of the calculator and obtain *x* =2.5.

Step 5. Apply the First Derivative Test. Enter *y*2=*d*(*y*1(*x* ), *x* ) and use the [*Zero*] function and obtain *x* =2.5. Note that:

Since f has only one relative extremum, it is the absolute extrenum.

Step 6. Thus at *x* =2.5, the length of the hypotenuse is the shortest. At The vertices of the triangle are (0, 0), (2.5, 0) and (0, 5).

- Rewrite:
*y*= [sin(cos(6*x*– 1))]^{2} - As
*x*→ ∞, the numerator approaches 0 and the denominator increases without bound (i.e., ∞). -
- Summarize the information of
*f '*on a number line. - The function f is decreasing on the interval (–∞, 3) and increasing on (3, ∞).
- The function
*f*is concave upward for the entire domain (–∞,∞). - Possible sketch of the graph for
*f*(*x*). See Figure 8.7-1.

Since

*f*has only one relative extremum, it is the absolute extremum. Thus, at*x*= 3, it is an absolute minimum.No change of concavity No point of inflection.

- Summarize the information of
- (Calculator) See Figure 8.7-2.
- (Calculator) See Figure 8.7-3.

Thus, =2 [sin (cos (6*x* – 1))]

× [cos(cos(6*x* – 1))]

× [–sin(6*x* – 1)] (6)

= – 12 sin(6*x* – 1)

× [sin(cos(6*x* – 1))]

× [cos(cos(6*x* – 1))].

Thus, the

Step 1. Differentiate:

Step 2. Set

Step 3. Solve for

Step 4. Thus,

Step 1. Distance formula:

Step 2. Enter:

Use the [Minimum] function of the calculator and obtain *x* = .65052 and *y*1 = .44488. Verify the result with the First Deriative Test. Enter *y*2 = *d*(*y*1 (*x*), *x* ) and use the [*Zero*] function and obtain *x* = .65052.

Thus the shortest distance is approximately 0.445.

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