Find the consumers' surplus and the producers' surplus at...

Find the consumers' surplus and the producers' surplus at the equilibrium price level for the given price-demand and price-supply equations. Include a graph that identifies the consumers' surplus and the producers' surplus. Round all values to the nearest integer.\ p = D(x) = 110e^{-0.001x}; p = S(x) = 30e^{0.001x}\ The value of x at equilibrium is\ (Round to the nearest whole number as needed.)

Transcript

00:01 Hi there, so for this problem we are asked to find the consumer surplus and the producer surplus at the equilibrium price level for the given price demand and price supplies equation.

00:13 Okay, we need to run all the values to the nearest integral, so the demand expression is 110 times the exponential of minus 0 .004 times x and the supply function will be 20 times the exponential of 0 .004 times x.

00:37 So let's determine first the value of x at equilibrium.

00:41 That value of x we determine by setting these two expressions equal to the others, so the only 110 times the exponential of minus 0 .004 times x is equal to 20 times the exponential of 0 .004 times x.

00:57 If we pass this to divide to the other side, we will have then that will be then, well and we can pass this to divide to the other side, so that will be 120 divided by 20, that will give us 11 divided by 2, then this, well that's a simplification, okay.

01:22 Then for the other side we'll have the exponential of 0 .008 times x.

01:28 Once we have this we need to solve for x, because that will be the value of x at equilibrium.

01:36 Right, so we obtain the neperian logarithm of 11 divided by 2, then this is equal to 0 .008 times x.

01:44 Finally we divide by 0 .08 in both sides, so let's use our calculator for this and the value that we obtain to the nearest integral is um 213.

02:21 Now the next question for this is the value of p at equilibrium, so we just evaluate the expression of p that we're given at this value, so that will be 110 times the exponential of minus 0 .004 times 213...

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