A boundary stripe 2 in. wide is painted around a rectangle whose dimensions are 100 ft by 240 ft. Use differentials to approximate the number of square feet of paint in the stripe.
A)113
B)113.33
C)113.81
D)113.89
E)113.23

Question

A boundary stripe 2 in. wide is painted around a rectangle whose dimensions are 100 ft by 240 ft. Use differentials to approximate the number of square feet of paint in the stripe.
A)113 
B)113.33 
C)113.81 
D)113.89 
E)113.23

Quizplus · Accepted Answer

The area of the rectangle is given by A = l*w = 100*240 = 24,000 sq ft. 
The length and width of the rectangle with the boundary stripe are (l+4) and (w+4), respectively. 
Therefore, the area of the rectangle with the boundary stripe is (l+4)*(w+4).
The area of the stripe is (l+4)*(w+4) - lw. 
Using differentials, we have: 
dA = (w+4)dl + (l+4)dw 
d(strip) = 2dl + 2dw

At (l,w)=(100,240), we have: 
dA = (240+4) * 1 + (100+4) * 1 = 248 
d(strip) = 2 * 1 + 2 * 1 = 4

So the approximate area of the stripe is: 
d(A_strip) = d(strip) * A + strip = 4 * 24,000 + strip

The length of the stripe is: 
l_strip = 2l + 2w + 8 = 2(100) + 2(240) + 8 = 688

The area of the stripe is: 
strip = l_strip * w_strip = 688 * 2 = 1376 sq ft

Therefore, 
d(A_strip) = 4 * 24,000 + 1376 = 25,376 sq ft

Rounding to two decimal places: 
d(A_strip) ≈ 113.33 sq ft

Therefore, the best choice is B.

A Boundary Stripe 2 In