The length of a rectangle is increasing at the rate of 2 feet per second, while the width is increasing at the rate of 1 foot per second. When the length is 5 feet and the width is 3 feet, how fast is the area increasing?
A) $5 \mathrm { ft } ^ { 2 } / \mathrm { s }$

B) $11 \mathrm { ft } ^ { 2 } / \mathrm { s }$
C) $6 \mathrm { ft } ^ { 2 } / \mathrm { s }$
D) $10 \mathrm { ft } ^ { 2 } / \mathrm { s }$
E) $15 \mathrm { ft } ^ { 2 } / \mathrm { s }$
F) $20 \mathrm { ft } ^ { 2 } / \mathrm { s }$

G) $8 \mathrm { ft } ^ { 2 } / \mathrm { s }$

H) $12 \mathrm { ft } ^ { 2 } / \mathrm { s }$

Question

The length of a rectangle is increasing at the rate of 2 feet per second, while the width is increasing at the rate of 1 foot per second. When the length is 5 feet and the width is 3 feet, how fast is the area increasing?
A) $5 \mathrm { ft } ^ { 2 } / \mathrm { s }$

B) $11 \mathrm { ft } ^ { 2 } / \mathrm { s }$
C) $6 \mathrm { ft } ^ { 2 } / \mathrm { s }$ 
D) $10 \mathrm { ft } ^ { 2 } / \mathrm { s }$
E) $15 \mathrm { ft } ^ { 2 } / \mathrm { s }$
F) $20 \mathrm { ft } ^ { 2 } / \mathrm { s }$

G) $8 \mathrm { ft } ^ { 2 } / \mathrm { s }$

H) $12 \mathrm { ft } ^ { 2 } / \mathrm { s }$

Quizplus · Accepted Answer

The area of a rectangle is given by the formula $A = lw$. The length is increasing at the rate of 2 feet per second, so $dl/dt = 2$. The width is increasing at the rate of 1 foot per second, so $dw/dt = 1$. When the length is 5 feet and the width is 3 feet, the area is $A = 5 \times 3 = 15$ square feet. The rate of change of the area is given by the product rule:$$\frac{dA}{dt} = \frac{dA}{dl} \frac{dl}{dt} + \frac{dA}{dw} \frac{dw}{dt}$$Substituting in the values we know, we get:$$\frac{dA}{dt} = (w)\frac{dl}{dt} + (l)\frac{dw}{dt}$$$$= (3)(2) + (5)(1)$$$$= 6 + 5$$$$= 11 \mathrm { ft } ^ { 2 } / \mathrm { s }$$

The Length of a Rectangle Is Increasing at the Rate 5ft2/s5 \mathrm { ft } ^ { 2 } / \mathrm { s }5ft2/s

The Length of a Rectangle Is Increasing at the Rate $5 \mathrm { ft } ^ { 2 } / \mathrm { s }$